GMAT 数学题（3）

sdcar2010

连续正整数如下图排列成一三角形。求列在第21行第一位的数值。

row 1							1
row 2						2	3	4
row 3					5	6	7	8	9
row 4				10	11	12	13	14	15	16

sdcar2010

Another solution
the number of digits in each row equals to   2*(the order number of the row)-1
so the number of digits in first 20 rows is
(1*2-1)+(2*2-1)+......+(20*2-1)
=(1+2+...+20)*2-20
=400
thus, the initial digit in 21st row is 401

-- by 会员 lonelyorchid (2011/1/2 14:59:52)

Good.

This is simply another way to express that
the number of number(s) on each row is an odd number:
number(s) in row n = 2*n-1

lonelyorchid

Another solution
the number of digits in each row equals to   2*(the order number of the row)-1
so the number of digits in first 20 rows is
(1*2-1)+(2*2-1)+......+(20*2-1)
=(1+2+...+20)*2-20
=400
thus, the initial digit in 21st row is 401

sdcar2010

牛人的对话。。。对我弹琴。。。

-- by 会员 gutou (2011/1/1 11:25:13)

Then you are TRUELY 牛!

gutou

牛人的对话。。。对我弹琴。。。

yayafl

刚才用了最苯的方法，列了数列。然后算的。401

第一次摸靠的时候就发现：有的题会做。但是就是无从下手。于是现在比较喜欢代入。呵呵
虽然比较慢。但是时间够用。

sdcar2010

ok
1+3+...+39 = 400
forgot to divide it by 2...

-- by 会员 gaor (2010/12/31 2:46:51)

You got it!

gaor

ok
1+3+...+39 = 400
forgot to divide it by 2...

sdcar2010

You got it!  The answer is 401. The last number on row n is n^2.  So the last number on row 20 is 400.  Then the first number on row 21 is 401.

落烟海

安规律发现：以1所在那一列为轴心的话
           第n行的轴心前后都有n-1项因此第n行有2*（n-1）+1项，即2n-1项
           而第n行的第一个数据an为一个等差数列相加：an=1+1+3+5+7+……+（2n-3）n>1          
所以第21行an=1+1+3+5+7+……+39=401不知道又没有算错，规律应该是没错的······