remainder play

[bblast]

find the greatest 4 digit number which when divided by 10,11,15,22 leaves 3,4,8,15 as remainder respectively.

a>9907
b>9903
c>9893
d>none

[jaymw]

Why are there only 4 answer choices, what is the source?

This question can be solved by process of elimination:

When you divide a number by 10, the units digit will be the remainder. Because the remainder for division by 10 is supposed to be 3, answer choice A can be eliminated.

Now try to divide B by 11. 900*11=9900, thus the remainder is 3. However, the remainder has to be 4 in this case.

We are down to C and D.

Because in B we already found out that 9900 is a multiple of 11, 9900-11=9889 will also be a multiple of 11. That means that the remainder of 9893/11 is 4.

This does not help us much, because the answer could still be C or D.

When dividing 9893 by 15, we get a remainder of 8. Still no help.

When dividing 9893 by 22, we get a remainder of 15. Bingo. [spoiler]C is the correct answer![/spoiler]

[manpsingh87]

find the greatest 4 digit number which when divided by 10,11,15,22 leaves 3,4,8,15 as remainder respectively.

a>9907
b>9903
c>9893
d>none

as none is one of the answer choices, therefore i believe to be on the safer side we must solve it instead of working on the answer choices...!!!

the difference between the divisors and their corresponding remainders is constant and is equal to 7; 10-3=11-7=15-8=22-15=7; also the hcf of the divisors 10,11,15,22=330;

now the largest four digit number is 9999; when we divide 9999 with 330 remainder is 99; therefore largest number that will divide 10,11,15,22=9900 hence our desired answer is 9900-7=9893 C

[bblast]

Why are there only 4 answer choices, what is the source?

This is a number properties question from a top Indian author. Not any exam directed but good stuff.

[jaymw]

Yep, I agree that it's a quality question. One more answer choice and it would make a good and rather hard GMAT problem!