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Introduction (赶时间的童鞋可以略过。。。只是一些概念帮助童鞋们回忆余数~~~)
Definition
If x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y= divisor * quotient + remainder = xq + r; and 0<=r < x.
For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since 15 = 6*2+3.
Notice that 0<= r < x means that remainder is a non-negative integer and always less than divisor.
This formula can also be written as y/x = q + r/x.
Properties
When y is divided by x the remainder is0 if y is a multiple of x.
For example, 12 divided by 3 yields the remainder of 0 since 12 is a multiple of 3 and 12 = 3*4+0.
When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer.
For example, 7 divided by 11 has the quotient 0 and the remainder 7 since 7=11*0+7
The possible remainders when positive integer y is divided by positive integer x can range from 0 to x-1.
For example, possible remainders when positive integer y is divided by 5 can range from 0 (when y is a multiple of 5) to 4 (when y is one less than a multiple of 5).
If a number is divided by 10, its remainder is the last digit of that number. If it is divided by 100 then the remainder is the last two digits and so on.
For example, 123 divided by 10 has the remainder 3 and 123 divided by 100 has the remainder of 23.
1. Collection of Methods
Method 1：小数法（妹纸自己取的名字，包括后面的方法也都是妹纸取的~欢迎讨论）
A way that the GMAT will test remainders is what you would typically just divide back into the problem to determine the decimals:
25/4 = 6 remainder 1
Divide that 1 back by 4 to get 0.25, so the answer is 6.25.
Any number with a remainder could be expressed as a decimal.
The remainder provides the data after the decimal point, and the quotient gives you the number to the left of the decimal point.
Consider this problem (which appears courtesy of GMAC):
Example： When positive integer x is divided by positive integer y, the remainder is 9. If x/y = 96.12,what is the value of y?
(A) 96 (B) 75 (C) 48 (D) 25 (E) 12
Sol：
Going back to the concept of the remainder, the remainder of 9 is what will give us that 0.12 after the decimal place. The answer to the division problem x/y is either:
96 remainder 9
Or
96.12
Therefore, when the remainder of 9 is divided back over y, we get 0.12. Mathematically, this means that:
9/y = 0.12
0.12y = 9
12y = 900
y = 900/12
y = 300/4
y = 75
The correct answer is B.
方法二：重建法
Given that an integer "n" when divided by an integer "a" gives "r" as reminder then the integer"n" can be written as
n = ak + r
where k is a constant integer.
Example 1： What is the remainder when B is divided by 6 if B is a positive integer?
(1) When B is divided by 18, the remainder is 3
(2) When B is divided by 12, the remainder is 9
Sol：
STAT1 : When B is divided by 18, the remainder is 3
So, we can write B as
B = 18k + 3
Now, to check the reminder when B is divided by 6, we essentially need to check the reminder when 18k + 3 is divided by 6
18k goes with 6 so the reminder will 3
So, it is sufficient
STAT2 : When B is divided by 12, the remainder is 9
So, we can write B as
B = 12k + 9
Now, to check the reminder when B is divided by 6, we essentially need to check the reminder when 12k + 9 is divided by 6
12k goes with 6 so the remainder will be the same as the reminder for 9 divided by 6 which is 3
So, reminder is 3
So, it is sufficient.
Answer will be D
Practice:
What is the remainder when positive integer t is divided by 5?
(1) When t is divided by 4, the remainder is 1
(2) When t is divided by 3, the remainder is 1
这题请大家自己试一试哦。毕竟嘛。。。看的效果还是不如自己做一题的效果好哇~~~下面大片的黄色区域都是讲解哦。。。所以大家不用担心~~~试着做一做吧~~~加油↖(^ω^)↗
Sol：
STAT1: When t is divided by 4, the remainder is 1
t = 4k +1
possible values of t are 1,5,9,13
Clearly we cannot find a unique reminder when t isdivided by 5 as in some cases(t=1) we are getting the reminder as 1 and insome(t=5) we are getting the reminder as 0.
So, INSUFFICIENT
STAT2: When t is divided by 3, the remainder is 1
t = 3s + 1
possible values of t are 1,4,7,10,13,16,19
Clearly we cannot find a unique reminder when t is divided by 5 as in some cases(t=1) we are getting the reminder as 1 and in some(t=10) we are getting the reminder as 0.
So, INSUFFICIENT
STAT1+STAT2： there are two approaches
1. Write the values of t from stat1 and then from stat2 and then take the common values
From STAT1 t = 1,5,9,13,17,21,25,29,33
From STAT2 t = 1,4,7,10,13,16,19,22,25,28,31,34
Common values are t = 1,13,25,
2. Equate t = 4k+1 to t=3s+1
We have 4k + 1 = 3s+1
k = 3s/4
since, k is an integer so only those values of s which are multiple of 4 will satisfy both STAT1 and STAT2
so, common values are given by t = 3s + 1 where s is multiple of 4
so t = 1,13,25 (for s=0,4,8 respectively)
Clearly we cannot find a unique reminder when t is divided by 5 as in some cases (t=1) we are getting the reminder as 1 and in some(t=10) we are getting the reminder as 0.
So, INSUFFICIENT
So, answer will be E
Example 2：If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?
(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1
Sol:
STAT1 : The remainder when p + n is divided by 5 is 1.
p+n = 5k + 1
but we cannot say anything about p^2 - n^2 just from this information.
So, INSUFFICIENT
STAT2 : The remainder when p - n is divided by 3 is 1
p-n = 3s + 1
but we cannot say anything about p^2 - n^2 just from this information.
So, INSUFFICIENT
STAT1+STAT2：
p^2 - n^2 = (p+n) * (p-n) = (5k + 1) * (3s + 1)= 15ks + 5k + 3s + 1
The reminder of the above expression by 15 is same as the reminder of 5k + 3s + 1 with 15 as 15ks will go with 15.
But we cannot say anything about the reminder as its value will change with the values of k and s.
So INSUFFICIENT
Hence answer will be E
Example 3：If n is a positive integer and r is the remainder when 4 +7n is divided by 3, what is the value of r?
(1) n+1 is divisible by 3
(2) n>20.
Sol:
r is the remainder when 4 + 7n is divided by 3
7n + 4 can we written as 6n + n + 3+ 1 = 3(2n+1) + n +1
reminder of 7n+4 by 3 will be same as reminder of 3(2n+1) +n +1 by 3
3*(2n+1) will go by 3 so the reminder will be the same as the reminder of (n+1) by 3.
STAT1: n+1 is divisible by 3
n+1 = 3k (where k is an integer)
n+1 will give 0 as the reminder when divided by 3
so, 7n+4 will also give 0 as the reminder when its divided by 3 (as its reminder is same as the reminder for (n+1) when divided by 3 => r =0
So, SUFFICIENT
STAT2: n>20.
we cannot do anything by this information as there are many values of n
so, INSUFFICIENT.
Hence, answer will be A
Practice: If x is an integer, is x between 27 and 54?
(1) The remainder when x is divided by 7 is 2.
(2) The remainder when x is divided by 3 is 2.
再次~~~请大家来考验下自己吧~~~O(∩_∩)O~~
Sol:
STAT1: The remainder when x is divided by 7 is 2.
x = 7k + 2
Possible values of x are 2,9,16,...,51,...
we cannot say anything about the values of x
so, INSUFFICIENT
STAT2: The remainder when x is divided by 3 is 2.
x = 3s + 2
Possible values of x are 2,5,8,11,...,53,...
we cannot say anything about the values of x
so, INSUFFICIENT
STAT1+STAT2：now there are two approaches
1. write the values of t from stat1 and then from stat2 and then take the common values
from STAT1 x = 2,9,16,23,30,37,44,51,58,...,65,...
from STAT2 x = 2,5,8,...,23,...,44,...,59,65,...
common values are x = 2,23,44,65,...
2. equate x = 7k+2 to x=3s+2
we have 7k + 2 = 3s+2
k = 3s/7
since, k is an integer so only those values of s which are multiple of 7 will satisfy both STAT1 and STAT2
so, common values are given by x = 3s + 2 where s is multiple of 7
so x = 2,23,44,65 (for s=0,7,14,21 respectively)
Clearly there are values of x which are between 27 and 54 (i.e. 44) and those which are not (2,23,65)
So, both together also INSUFFICIENT
So, answer will be E
方法三：MOD法
请大家参见知之为之之大侠的帖子！！！地址：<http://forum.chasedream.com/GMAT_Math/thread-437516-1-1.html>
2. Remainder QuestionPatterns
Background
l Most GMAT remainder problems are encountered in data sufficiency section.
l All GMAT remainder questions are limited to positive integers only.
l Both number plugging method and algebra are suitable tosolve remainder questions.
l Some remainder questions can be disguised as word problems. See below.
l Usually you get 1, maximum 2 questions on remainders on thetest (based on GMAT Prep CATs)
以下Pattern并不是按照重要性顺序来排的哟~~~Pattern 6 is the most common pattern! 妹纸觉得大家都可以看一看。。。当做练习吧~~~
Pattern#1: The ratio of two integers is given and we are asked to find possible value of the remainder when one integer is divided by another.
Q1： OG13 diagnostic test, question 13
If s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t ?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45
Sol：
S divided by t yields the remainder of r can always be expressed as: s/t = q + r/t (which is the same as s= qt+r), where q is the quotient and r is the remainder.
Given that s/t = 64.12 = 64(12/100) = 64(3/25) = 64 +3/25,so according to the above r/t=3/25, which means that r must be a multiple of 3. Only option E offers answer which is a multiple of 3
Answer. E.
Q2： OG13 Practice Questions, question 95
When positive integer x is divided by positive integer y,the remainder is 9. If x/y = 96.12, what is the value of y?
(A) 96
(B) 75
(C) 48
(D) 25
(E) 12
Sol：
When positive integer x is divided by positive integer y,the remainder is 9 --> x=qy+9;
x/y=96.12 --> x=96y+0.12y (so q above equals to 96);
0.12y=9 --> y=75.
Answer: B
finding the remainder when an expression with variable is divided by some integer.
OG13 Practice Questions, question 26
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12 ?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 5
Sol：
There are several algebraic ways to solve this question, but the easiest way is as follows: since we cannot have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12.
n=5 --> n^2=25 --> remainder upon division 25 by 12 is 1.
Answer: B.
Pattern#3: min/max question involving remainders
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35.
A. 3
B. 4
C. 12
D. 32
E. 35
Sol：
Positive integer n is divided by 5, the remainder is 1--> n = 5q+1, where q is the quotient --> 1, 6, 11, 16,21, 26,31, ...
Positive integer n is divided by 7, the remainder is 3--> n=7p+3, where p is the quotient --> 3, 10, 17, 24, 31,....
You cannot use the same variable for quotients in both formulas, because quotient may not be the same upon division n by two different numbers.
For example 31/5, quotient q=6 but 31/7, quotient p=4.
There is a way to derive general formula for n (of a type n= mx+r, where x is divisor and r is a remainder)based on above two statements:
Divisor x would be the least common multiple of above two divisors 5 and 7, hence x=35.
Remainder r would be the first common integer in above two patterns, hence r=31.
Therefore general formula based on both statements is n=35m+31.Thus the smallest positive integer k such that k+n is a multiple of 35 is 4--> n+4 = 35k+31+4 = 35(k+1).
Answer: B
Pattern#4: disguised PS remainder problem.
There are between 100 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection?
(A) 101
(B) 103
(C) 106
(D) 107
(E) 109
Sol：
If the cards are counted out 3 at a time, there are 2 leftover: x=3q+2. From the numbers from 100 to 110 following three give the remainder of 2 upon division by 3: 101, 104 and 107;
If the cards are counted out 4 at a time, there are 1 leftover: x=4p+1. From the numbers from 100 to 110 following three give the remainder of 1 upon division by 4: 101, 105 and 109.
Since x, the number of cards, should satisfy both conditions then it equals to 101.
Answer:A
Pattern#5: we need to answer some question about an integer, when the statements give info involving remainders.
Q1：OG13 Practice Questions, question 58
What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30.
(2) x divided by 110 has a remainder of 30.
Sol：
(1) x divided by 100 has a remainder of 30 --> x =100q+30, so x can be: 30, 130, 230, ... Each has the tens digit of 3. Sufficient.
(2) x divided by 110 has a remainder of 30 --> x =110p+30, so x can be: 30, 250, ... We already have two values for the tens digit. Not sufficient.
Answer:A.
Q2：OG13 Practice Questions, question 83
If k is an integer such that 56 < k < 66, what is the value of k ?
(1) If k were divided by 2, the remainder would be 1.
(2) If k + 1 were divided by 3, the remainder would be 0.
Sol：
(1) If k were divided by 2, the remainder would be 1 -->k is an odd number, thus it could be 57, 59, 61, 63, or 65. Not sufficient.
(2) If k + 1 were divided by 3, the remainder would be 0--> k is 1 less than a multiple of 3, thus it could be 59, 62, or 65. Not sufficient.
(1)+(2) k could still take more than one value: 59 or 65. Not sufficient.
Answer: E.
Pattern#6: we need to find the remainder when some variable or an expression with variable(s) is divided by some integer. Usually the statements give divisibility/remainder info. Most common patter.
Q1：What is the remainder when the positive integer n is divided by 6?
(1) n is multiple of 5
(2) n is a multiple of 12
Sol：
(1) n is multiple of 5. If n=5, then n yields the remainder of 5 when divided by 6 but if n=10, then n yields the remainder of 4 when divided by 6. We already have two different answers, which means that this statement is not sufficient.
(2) n is a multiple of 12. Every multiple of 12 is also a multiple of 6, thus n divided by 6 yields the remainder of 0. Sufficient.
Answer: B.
Q2：If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?
(1) When x-y is divided by 5, the remainder is 1
(2) When x+y is divided by 5, the remainder is 2
Sol：
(1) When x-y is divided by 5, the remainder is 1--> x-y = 5q+1, so x-y can be 1, 6, 11, ... Now, x=2 and y=1 (x-y=1) then x^2+y^2= 5 and thus the remainder is 0, but if x=3 and y=2 (x-y=1)then x^2+y^2= 13 and thus the remainder is 3. Not sufficient.
(2) When x+y is divided by 5, the remainder is 2--> x+y=5p+2, so x+y can be 2, 7, 12, ... Now, x=1 and y=1 (x+y=2) then x^2+y^2= 2 and thus the remainder is 2, but if x=5 and y=2 (x+y=7) then x^2+y^2= 29 and thus the remainder is 4. Not sufficient.
(1)+(2) Square both expressions: x^2 -2xy +y^2= 25q^2 + 10q+ 1 and x^2 +2xy +y^2= 25p^2 + 20p + 4 --> add them up: 2(x^2+y^2)= 5(5q^2+2q+5p^2+4p+1)--> so 2(x^2+y^2) is divisible by 5 (remainder 0), which means that so is x^2+y^2. Sufficient.
Answer: C.
Q3：If t is a positive integer and r is the remainder when t^2+5t+6 is divided by 7, what is the value of r?
(1) When t is divided by 7, the remainder is 6.
(2) When t^2 is divided by 7, the remainder is 1.
Sol：
First of all factor t^2+5t+6 --> t^2+5t+6=(t+2)(t+3)
(1) When t is divided by 7, the remainder is 6 -->t=7q+6 --> (t+2)(t+3)=(7q+8)(7q+9). Now, no need to expand and multiply all the terms, just notice that when we expand all terms but the last one, which will be 8*9=72, will have 7 as a factor and 72 yields the remainder of 2 upon division by 7. Sufficient.
(2) When t^2 is divided by 7, the remainder is 1 -->different values of t possible: for example t=1 or t=6, which when substituted in (t+2)(t+3) will yield different remainder upon division by 7. Not sufficient.
Answer: A.
Q4：If p is a positive odd integer, what is the remainder when p is divided by 4 ?:
(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.
Sol：
(1) When p is divided by 8, the remainder is 5 --> p= 8q+5 = (8q+4) + 1 = 4(2q+1) + 1--> so the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.
(2) p is the sum of the squares of two positive integers -->since p is an odd integer then one of the integers must be even and another odd: p=(2n)^2 + (2m+1)^2 = 4n^2 + 4m^2 + 4m + 1 = 4(n^2+m^2+m) + 1-->the same way as above: the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.
Answer: D.
Q5：If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?
(1) n is not divisible by 2
(2) n is not divisible by 3
Sol：
Plug-in method:
(n-1)(n+1) = n^2-1
(1) n is not divisible by 2 --> pick two odd numbers:let's say 1 and 3 --> if n=1, then n^2-1=0 and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if n=3, then n^2-1=8 and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.
(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if n=1, then n^2-1=0,so remainder is 0 but if n=2, then n^2-1=3 and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.
(1)+(2) Let's check for several numbers which are not divisible by 2 or 3:
n=1 --> n^2-1=0 --> remainder 0;
n=5 --> n^2-1=24 --> remainder 0;
n=7 --> n^2-1=48 --> remainder 0;
n=11 --> n^2-1=120 --> remainder 0.
Well it seems that all appropriate numbers will give remainder of 0. Sufficient.
Algebraic approach:
(1) n is not divisible by 2. Insufficient on its own, but this statement says that n=odd --> n-1 and n+1 are consecutive even integers --> (n-1)(n+1) must be divisible by 8(as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8,10); (10, 12), ...).
(2) n is not divisible by 3. Insufficient on its own, but form this statement either n-1 or n+1 must be divisible by 3 (as n-1, n, and n+1 are consecutive integers, so one of them must be divisible by 3, we are told that it's not n, hence either n-1 or n+1).
(1)+(2) From (1) (n-1)(n+1) is divisible by 8,from (2) it's also divisible by 3, therefore it must be divisible by 8*3=24,which means that remainder upon division (n-1)(n+1) by 24 will be 0. Sufficient.
Answer: C.
Pattern#7: disguised DS remainder problem.
A person inherited few gold coins from his father. If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over. What is the number of coins he inherited from his father?
(1) The number of coins lies between 50 to 120.
(2) If he put 13 coins in one bag then no coin is left over and number of coins being lesser than 200.
Sol：
If he puts 9 coins in each bag then 7 coins are left over--> c=9q+7, so # of coins can be: 7, 16, 25, 34, 43, 52, 61, ...
If he puts 7 coins in each bag then 3 coins are left over--> c=7p+3, so # of coins can be: 3, 10, 17, 24, 31, 38, 45,52, 59, ...
General formula for c based on above two statements will be: c=63k+52 (the divisor should be the least common multiple of above two divisors 9 and 7, so 63 and the remainder should be the first common integer in above two patterns, hence 52).
C=63k+52 means that # of coins can be: 52, 115, 178,241, ...
(1) The number of coins lies between 50 to 120 --> # of coins can be 52 or 115. Not sufficient.
(2) If he put 13 coins in one bag then no coin is left over and number of coins being lesser than 200 --> # of coins is a multiple of 13 and less than 200: only 52 satisfies this condition. Sufficient.
Answer: B.
Pattern#8: C-Trap remainder problem. "C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer.
If a and b are positive integers, what is the remainder when 4^(2a+1+b) is divided by 10?
(1) a = 1
(2) b = 2
Sol：
4 in positive integer power can have only 2 last digits: 4,when the power is odd or 6 when the power is even. Hence, to get the remainder of 4^x/10 we should know whether the power is odd or even: if it's odd the remainder will be 4 and if it's even the remainder will be 6.
(1) a = 1 --> 4^(2a+1+b) = 4^(3+b) depending on b the power can be even or odd. Not sufficient.
(2) b = 2 --> 4^(2a+1+b) = 4^(2a+3) = 4^(even+odd)= 4^odd--> the remainder upon division of 4^odd by 10 is 4. Sufficient.
Answer: B.
3. Collection of GMATRemainder Problems
这些题目妹纸觉得还是自己做一遍会很有用哦！！！所以妹纸把分析全部隐藏在黄色highlight里面啦。。。大家加油~~~~
1.If r is the remainder when the positive integer n is divided by 7, what is the value of r
(1) When n is divided by 21, the remainder is an odd number
(2) When n is divided by 28, the remainder is 3
Sol：
The possible reminders can be 1,2,3,4,5 and 6. We have the pinpoint the exact remainder from this 6 numbers.
St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number.
But it cannot be 7, 3 or 9 . Hence the possibilities are: 1 and 5.
Hence there can be two remainders ,1 and 5, when divided by 7.
NOT SUFFICIENT
St 2: when n is divided by 28 the remainder is 3.
As 7 is a factor of 28, the remainder when divided by 7 will be 3
SUFFICIENT
Answer: B
2. If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?
(1) n = 2
(2) m = 1
Sol：
The Concept tested here is cycles of powers of 3.
The cycles of powers of 3 are : 3,9,7,1
St I) n = 2. This makes 3^(4*2 +2 + m) = 3^(10+m). we do not know m and hence cannot figure out the unit digit.
St II) m=1 . This makes 3^(4*n +2 + 1).
4n can be 4,8,12,16...
3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7. SUFF
Hence B
3.If p is a positive odd integer, what is the remainder when p is divided by 4 ?
(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.
Sol：
st1. take multiples of 8....divide them by 4...remainder=1 in each case...
st2. p is odd, since p is square of 2 integers...one will be even and other odd....now when we divide any even square by 4 vll gt 0 remainder..and when divide odd square vll get 1 as remainder......so in total remainder=1
Ans : D
4.If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?
(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.
Sol：
st1) p+n=6,11,16....insuff.
st2) p-n=4,7,10....insuff...
multiply these two to get p^2-n^2.....multiplying any two values from the above results in different remainder......
also can be done thru equation....p+n=5a+1..and so on
Ans: E
5.What is the remainder when the positive integer x is divided by 3 ?
(1) When x is divided by 6, the remainder is 2.
(2) When x is divided by 15, the remainder is 2.
Sol：
st 1：multiple of 6 will also be multiple of 3 so remainder will be same as 2.
st 2：multiple of 15 will also be multiple of 3....so the no.that gives remainder 2 when divided by 15 also gives 2 as the remainder when divided by 3...
Answer D
6.What is the remainder when the positive integer n is divided by 6 ?
(1) n is a multiple of 5.
(2) n is a multiple of 12.
Sol：
st 1) multiples of 5=5,10,15....all gives differnt remainders with 6
st 2) n is divided by 12...so it will be divided by 6...remainder=0
Answer B
7. If x, y, and z are positive integers, what is the remainder when 100x + 10y + z is divided by 7?
(1) y = 6
(2) z = 3
Sol：
We need to know all the variables. We cannot get that from both the statements. Hence the answer is E.
8. If n is a positive integer and r is the remainder when 4 +7n is divided by 3, what is the value of r ?
(1) n + 1 is divisible by 3.
(2) n > 20
Sol：
st1) n+1 divisible by 3..so n=2,5,8,11......
this gives 4+7n=18,39,60....remainder 0 in each case......
st2) insufficient ....n can have any value
Answer A
9. If n is a positive integer and r is the remainder when (n- 1)(n + 1) is divided by 24, what is the value of r ?
(1) n is not divisible by 2.
(2) n is not divisible by 3.
Sol：
ST 1- if n is not divisible by 2, then n is odd, so both(n - 1) and (n + 1) are even. Moreover, since every other even number is a multiple of 4, one of those two factors is a multiple of 4. So the product (n -1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x 2 x 2 = three 2's in its prime factorization.
But this is not sufficient, because it can be (n-1)*(n+1)can be 2*4 where remainder is 8. it can be 4*6 where the remainder is 0.
ST 2- if n is not divisible by 3, then exactly one of (n- 1) and (n + 1) is divisible by 3, because every third integer is divisible by 3. Therefore, the product (n - 1)(n + 1) contains a 3 in its prime factorization.
Just like st 1 this is not sufficient
ST1+ST2 - the overall prime factorization of (n - 1)(n +1) contains three 2's and a 3.
Therefore, it is a multiple of 24.
Sufficient
Answer C |
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