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remainder problem

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remainder problem

by clar » Tue Aug 18, 2009 12:22 am
When n is divided by 5 the remainder is 2. When n is divided by 7 the remainder is 1. If n is less than 800 , how many values of n are possible?

a)22
b)23
c)24
d)25
e)26
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by tohellandback » Tue Aug 18, 2009 12:47 am
IMO B
number is of the form
5k+2 and 7n+1
first number is 22
add the multiples of the LCM of 5 and 7 and you get the series

22, 22+35, 22+2*35...., 792

total numbers: 792-22/35 +1
22+1=23
The powers of two are bloody impolite!!
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by prindaroy » Tue Aug 18, 2009 12:00 pm
Another approach:

n = 5p + 2

n = 7q + 1

5p + 2 = 7q + 1

5p + 1 = 7q, so for all numbers of q where the number is a multiple of 5 plus 1. Additionally remember for such numbers that the end has to be either 1 or 6 in order for it to be of the form 5p + 1, since all multiples of 5 end in 0 or 5. So to create a units digit of 1 or 6, 7 needs to be multiplied by 3 or 8.

so;

we have 7*3, 7*13,7*13.......7*113, count those and you get 12

7*8,7*18,.......,7*108, count those and you get 11. Add the two and the answer is 23.
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