大菜分享个题目

spicytaco · 发表于 2018-10-21 12:11:46

本人数学都忘光了，所以什么都要重头学，看到个题目觉得挺不错的，分享给大家

If n is an integer greater than 1, is 3^n-2^n divisible by 35?

(1) n is divisible by 15.
(2) n is divisible by 18.

解析：

RULE: for x^n-y^n:

X^n−y^n is ALWAYS divisible by x−y.
x^n−y^n is divisible by x+y when n is even

If n is an integer >1, is 3n−2n3n−2n divisible by 35?

(1) 3^15n-2^15n = (3^3)^5n - (2^3)^5n = 27^5n - 8^5n

considering n to be 1 (odd) we get that 27^5 ends with a 7, and 8^5 ends with an 8. so the last digit will be 9 .

35 cant divide a number ending with 9

consider n to be even now, say 2 i.e. 27^10 - 8^10 .

now we will need some corollaries we get from the binomail theorem. if we have a number as a^n - b^n

then if n is odd , the number is divisible by a - b
if n is even, the number is divisible by a + b

so this makes 27^10-8^10 divisible by 27 + 8 = 35.

thus an even value of n will make the term 3^15n - 2^15n divisible by 35, while an odd value will not be divisble

So 1 is insufficient.

By using the above mentioned rule we can infer that 3^18n-2^18n (27^6n - 8 ^6n) will always be divisible by 35 as the index will always be even.

So 2 is sufficient .

artemis_ys · 发表于 2018-10-21 12:30:17

好棒，学到了

大菜分享个题目

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