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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

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

You should also notice that N only satisfies 5 if it ends with 2 or 7. Since we know n starts with 22 we can just keep adding 7s to it (to keep 7's remainder) until we hit a number that ends in 2 or 7 to satisfy 5 (7s already satisfied since we kept adding 7s to it). after getting the next one you can then divide it out and find how many "35 number gaps" there are. _________________

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

I see how u get 22, 57 (most likely just by guessing) but I am not able to follow your logic how u find the formula, where u multiply 35 by 22, and get 770 adding 22 u get 23Ns? Can u explain please in detail? Thank u _________________

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

I see how u get 22, 57 (most likely just by guessing) but I am not able to follow your logic how u find the formula, where u multiply 35 by 22, and get 770 adding 22 u get 23Ns? Can u explain please in detail? Thank u

well we start with 22. Now we need to find the next number that ends in 2 or 7 to satisfy 5's criteria. Since 22 already has remainder of 1 for 7 all we need to do is add 7 to it to find the next number ends in 2 or 7. After finding that number (57) we see the gap is 35. it will keep being 35 since the remainder will keep switching like before until the next N is found. since we know the "gap" is 35 then we multiply by 22( the lowest possible answer) to get 770. Since we know the value starts at 22 and 792 is only a few behind 800 so we stop there. You could also see it as : 22 + 35*x = 800 and find the value that way but I was lazy and used the answer choices (plus it is pretty easy to do 35*22 on the fly). It is probably no coincidence that the gap is 5 X 7 (both values in question are in it) but the number property escapes me so i dont want to feed you the wrong info. But even without it you can get to the 2nd value pretty quick and notice the pattern (can add another 35 to it to double check i guess...). _________________

I think you are right. It is not a coincidence that you add 35, which is LCM(5,7). I think the following might explain why LCM should be the gap between the numbers. We have two equations: n=5X+2 n=7y+1 To find the next number, so that n=5*integer+2, the multiple of 5 should be added to the previous number: n+5k=5x+5k+2=5*(x+k)+2. So, the new number still has a remainder of 2 when divided by 5. By the same logic, a multiple of 7 should be added to leave a remainder of 1 when divided by 7. It seems to me that if you add a multiple of a divisor to a dividend, a remainder is unchanged when a new number (dividend plus a multiple of divisor) is divided by the same divisor. To leave a remainder of 2 when divided by 5 and a remainder of 1 when divided by 7, a multiple of both 5 and 7 should be added. Since we are asked to find the maximum, we should add the smallest multiple - LCM: n1=22 n2=n1+35 n3=n2+35 etc...

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

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

You should also notice that N only satisfies 5 if it ends with 2 or 7. Since we know n starts with 22 we can just keep adding 7s to it (to keep 7's remainder) until we hit a number that ends in 2 or 7 to satisfy 5 (7s already satisfied since we kept adding 7s to it). after getting the next one you can then divide it out and find how many "35 number gaps" there are.

I came up with 35 as well after writing it out. No way I could ever do this in 2min.

Interestingly after coming up with 35... and plowing through to the answer I get 23. An easier way... that may work (not sure whether its full proof). 800/35= 22.857=~23...

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

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

You should also notice that N only satisfies 5 if it ends with 2 or 7. Since we know n starts with 22 we can just keep adding 7s to it (to keep 7's remainder) until we hit a number that ends in 2 or 7 to satisfy 5 (7s already satisfied since we kept adding 7s to it). after getting the next one you can then divide it out and find how many "35 number gaps" there are.

I came up with 35 as well after writing it out. No way I could ever do this in 2min.

Interestingly after coming up with 35... and plowing through to the answer I get 23. An easier way... that may work (not sure whether its full proof). 800/35= 22.857=~23...

I dont think you can take it approx 23 ..it has to be 22...... The answer is 23 because it adds the first value ie 22 which the calculation doesnt include.

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

is it B? I see that the first N is 22 that satisfies the criteria. Then 35 values later 57 satisfies and so on... multiplying 35 X 22 we get 770 and adding the 22 from the beginning we have 23 Ns

You should also notice that N only satisfies 5 if it ends with 2 or 7. Since we know n starts with 22 we can just keep adding 7s to it (to keep 7's remainder) until we hit a number that ends in 2 or 7 to satisfy 5 (7s already satisfied since we kept adding 7s to it). after getting the next one you can then divide it out and find how many "35 number gaps" there are.

I came up with 35 as well after writing it out. No way I could ever do this in 2min.

Interestingly after coming up with 35... and plowing through to the answer I get 23. An easier way... that may work (not sure whether its full proof). 800/35= 22.857=~23...

You can come up with 35 very fast. The difference between the numbers should be LCM (least common multiple) of 5 and 7. Otherwise n will not satisfy both equations : n=5*x+2 and n=7*y+1. As soon as I got 35, I figured out that the first number is 22 (simply by picking numbers). Then (800-22)/35 will be 22.22. So there should be 22 numbers not including the first number and 23 in total.

Thus, T (number of Ns) is equal to 22, because the nearest integer, which is less than 778/35 is 22. However, if we insert the T (22) in a formula N=35T+22, we can see that there are 23 integers on the number line, which is least than 800.

Excuse me for my previous mistake. Definitely B, 23 _________________

Never, never, never give up

Last edited by barakhaiev on 02 Sep 2009, 11:51, edited 1 time in total.

Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
27 Feb 2012, 13:24

4

This post received KUDOS

Expert's post

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yezz wrote:

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

First of all it should be mentioned in the stem that \(n\) is a positive integer: every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

Given: \(n=5p+2\), so \(n\) can be: 2, 7, 12, 17, 22, 27, ... \(n=7q+1\), so \(n\) can be: 1, 8, 15, 22, 29, ...

General formula for \(n\) based on above two statements will be: \(n=35k+22\) (the divisor should be the least common multiple of above two divisors 5 and 7, so 35 and the remainder should be the first common integer in above two patterns, hence 22).

Next, given that \(35k+22<800\) --> \(k<22.something\) --> so \(k\), can take 23 values from 0 to 22, inclusive, which means that \(n\) also can take 23 values (the least value of \(n\) will be 22 for \(k=0\)).

Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
09 May 2012, 01:06

Bunuel wrote:

yezz wrote:

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

First of all it should be mentioned in the stem that \(n\) is a positive integer: every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

Given: \(n=5p+2\), so \(n\) can be: 2, 7, 12, 17, 22, 27, ... \(n=7q+1\), so \(n\) can be: 1, 8, 15, 22, 29, ...

General formula for \(n\) based on above two statements will be: \(n=35k+22\) (the divisor should be the least common multiple of above two divisors 5 and 7, so 35 and the remainder should be the first common integer in above two patterns, hence 22).

Next, given that \(35k+22<800\) --> \(k<22.something\) --> so \(k\), can take 23 values from 0 to 22, inclusive, which means that \(n\) also can take 23 values (the least value of \(n\) will be 22 for \(k=0\)).

Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
09 May 2012, 01:10

Expert's post

keiraria wrote:

Bunuel wrote:

yezz wrote:

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

First of all it should be mentioned in the stem that \(n\) is a positive integer: every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

Given: \(n=5p+2\), so \(n\) can be: 2, 7, 12, 17, 22, 27, ... \(n=7q+1\), so \(n\) can be: 1, 8, 15, 22, 29, ...

General formula for \(n\) based on above two statements will be: \(n=35k+22\) (the divisor should be the least common multiple of above two divisors 5 and 7, so 35 and the remainder should be the first common integer in above two patterns, hence 22).

Next, given that \(35k+22<800\) --> \(k<22.something\) --> so \(k\), can take 23 values from 0 to 22, inclusive, which means that \(n\) also can take 23 values (the least value of \(n\) will be 22 for \(k=0\)).

Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
09 May 2012, 04:16

Quote:

General formula for n based on above two statements will be: n=35k+22 (the divisor should be the least common multiple of above two divisors 5 and 7, so 35 and the remainder should be the first common integer in above two patterns, hence 22).

I don't understand why the remainder should be the first common integer (22).. could you please explain?

Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
09 May 2012, 05:53

1

This post received KUDOS

Expert's post

jacket882 wrote:

Quote:

General formula for n based on above two statements will be: n=35k+22 (the divisor should be the least common multiple of above two divisors 5 and 7, so 35 and the remainder should be the first common integer in above two patterns, hence 22).

I don't understand why the remainder should be the first common integer (22).. could you please explain?

22 is the first number that satisfies both conditions, so in general formula it should be the first possible value of n (for k=0). _________________

Re: When n is divided by 5 the remainder is 2. When n is divided [#permalink]
14 Jul 2012, 01:48

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?

n = 5x +2, so the unit place must be 2 or 7 n = 7y +1 --> n-1 = 7y, so (n-1) is divisible by 7 or I need to sort out all the numbers which are divisible by 7 with the unit place of 1 or 6.

Look at the first 10 multipications of 7, 7*1 = 7 7*2 = 14 7*3 = 21 7*4 = 28 etc. we have the unit place are 7, 4, 1, 8, 5, 2, 9, 6, 3, 0, respectively. This round will repeat endlessly. Obviously, we can pick 2 out of those first 70 numbers which satisfy the 2 requirements above (divisible by 7 and unit place is either 1 or 6). From 0 to 799 we have 11 rounds like that, that means we can pick 11*2=22 numbers out of 770 numbers. From 771 to 799 we don't have an entire round, but from 7, 4, 1, 8, and we can pick 1 number more.

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