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Tough PS-2

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Tough PS-2

by satishchandra » Sat Oct 27, 2012 6:37 am
Find the sum of all three digit numbers which leave a remainder 2 when divided by either 7 or 5.
(A) 14179 (B) 14157 (C) 15011 (D) 14171 (E) 14000
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by pemdas » Sat Oct 27, 2012 8:35 am
satishchandra wrote:Find the sum of all three digit numbers which leave a remainder 2 when divided by either 7 or 5.
(A) 14179 (B) 14157 (C) 15011 (D) 14171 (E) 14000
abc=N

N=7x + 2 = 5y + 2, 7x = 5y where both x and y are integers. Since both 5 and 7 are primes their products should get themselves to be equal, i.e. x=5 and y=7.

N=7*5+2=37 means that our three digit number abc should be a factor of 35 and leaving a remainder of 2 always. This is achieved by factoring 35 onto 3,4,5 all the way through 28 (980/35). So we have (28-3+1=26) factors of 35, implying we must have 26 numbers.

These 26 numbers' sum will be 35*(3+4+5... 28) or 35*(28+3)*26/2=14105 ADDED by 2*26=52 gives 14157.

answer b)
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by abhi131 » Wed Oct 31, 2012 6:09 pm
Basing partly on the previous answer,
abc should be a factor of 35 and leaving a remainder of 2 always.
we know that the difference between consecutive terms will be 35. Let's now find the first and last terms.
Mental calculation yields - first term is 35*3+2 = 107.
For the last term, divide 999 by 35, gets a remainder of 19. To get a remainder of 2, last term should be 999-17 = 982.
So we know,
first term, last term, and difference. Number of terms = (last term- first term)/2 + 1 = 26.
Sum = Number of terms*(first term + last term)/2 = 26(982+107)/2 = 14157.[/u]
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