Math, asked by hasini24072006, 5 hours ago

let p be the smallest positive integer n for which the last 3 digits of 2007n are 837 Find the value of p-1/30

Answers

Answered by audhityamohan
4

Answer:

the answer's 38

Answered by amitnrw
10

Given :  2007 * n

To Find : smallest positive integer n for which the last three digits of 2007 * n are 837

value of (n-1)/30

Solution:

2007 * n  =  _______837

hence unit digit of n must be 1

as  7 * 1 = 7     is only possible

Hence 2007 * 1 = 2007

Hence 200 is carry over

so  200  +  2007 * x   must end with 3

and only possible value of x = 9

so tens digit of n is  9

200 + 2007* 9 =  18263

Hence 1826  is carry over

1826  + 2007 * y   must end with  8

=> hence 2007 * y  must end with 2

=> y  = 6  

Hence 100 digit of n = 6

So 691  is  smallest   positive integer n for which the last three digits of 2007n are 837

(691 - 1)/30 =  23

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