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The three digit positive integer N = a2b is both a multiple

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The three digit positive integer N = a2b is both a multiple [#permalink] New post  Updated on: 25 Feb 2014, 07:25
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The three digit positive integer N = a2b is both a multiple of 3 and a multiple of 5. What is N?

(1) a + b is an even integer
(2) a/b = 1
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Originally posted by gmattokyo on 01 Nov 2009, 04:11.
Last edited by Bunuel on 25 Feb 2014, 07:25, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: N=a2b. What is N? [#permalink] New post  01 Nov 2009, 04:34
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The three digit positive integer N = a2b is both a multiple of 3 and a multiple of 5. What is N?

N=a2b
N is a multiple of 3 --> a + 2 + b = {a multiple of 3}.
N is a multiple of 5 --> b = 0 or b = 5.

If b = 0, then a can be 1, 4, or 7 (as a+2+b must be multiple of 3).
If b = 5, then a can be 2, 5, or 8 (for the same reason).

Total of 6 pairs of a and b thus 6 possible values of N.

(1) a + b is an even integer --> two pairs satisfy this b=0, a=4: N=420 OR b=5, a=5: N=525. Not sufficient.

(2) a/b = 1 --> only one possible combination for a and b: a=5, b=5 --> N=525. Sufficient.

Answer: B.
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Re: N=a2b. What is N? [#permalink] New post  01 Nov 2009, 04:51
another bingo!
I think it's the dagger in your hand. Helps chop at these dodgy problems ;)
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Re: N=a2b. What is N? [#permalink] New post  01 Nov 2009, 07:53
Statement 1) Insufficient
b must be 0 or 5, multiple possibilities for a (a+2+b must be divisible by 3)

Statement 2) Sufficient
Only one possibility a=b=5 (As a and b can not be 0). So N is 525.

Answer is B.
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Re: N=a2b. What is N? [#permalink] New post  06 Nov 2009, 10:04
Another way to solve it

since the number is both multiple of 3 and 5, therefore the Least common multiple would be 15 i.e. a2b has to be a multiple of 15.

so possible 2 digits numbers, that are multiple of 15, with 2 in tens place = {120, 225, 420, 525, 720, 825}

Statement 1: a + b is even; not sufficient; as with both 420 and 525 "a + b" is even
Statement 2: a/b=1; sufficient; as only 525 has a = b

Therefore answer is B
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Re: The three digit positive integer N = a2b is both a multiple [#permalink] New post  25 Feb 2014, 07:30
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Bumping for review and further discussion.
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Re: The three digit positive integer N = a2b is both a multiple [#permalink] New post  26 Feb 2014, 02:13
gmattokyo wrote:
The three digit positive integer N = a2b is both a multiple of 3 and a multiple of 5. What is N?

(1) a + b is an even integer
(2) a/b = 1



My take is B

for a2b to be a multiple of 5, it can be:

(i) a25 or (ii) a20

for a25 to be a multiple of 3, it can be --> 120, 420, 720.
for a20 to be a multiple of 3, it can be --> 225, 525, 825.

stmnt 1) a+b = even
here 420 and 525 both hold true. (so insufficient)

stmnt2) a/b = 1
here only 525 holds. (hence 2 alone is sufficient)
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Re: N=a2b. What is N? [#permalink] New post  01 Mar 2014, 04:03
Bunuel wrote:
The three digit positive integer N = a2b is both a multiple of 3 and a multiple of 5. What is N?

N=a2b
N is a multiple of 3 --> a + 2 + b = {a multiple of 3}.
N is a multiple of 5 --> b = 0 or b = 5.

If b = 0, then a can be 1, 4, or 7 (as a+2+b must be multiple of 3).
If b = 5, then a can be 2, 5, or 8 (for the same reason).

Total of 6 pairs of a and b thus 6 possible values of N.

(1) a + b is an even integer --> two pairs satisfy this b=0, a=4: N=420 OR b=5, a=5: N=525. Not sufficient.

(2) a/b = 1 --> only one possible combination for a and b: a=5, b=5 --> N=525. Sufficient.

Answer: B.


I dont understand how is that you have deduced that b must be a 0 or 5. I realize that any multiple of 5 has to have for its last digit either a 0 or 5. But here N = a2b is simply N = (a) (2) (b) and thus could have been written in a number of other ways. For example, N = 2ba.
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Re: N=a2b. What is N? [#permalink] New post  01 Mar 2014, 04:07
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purijasv123 wrote:
Bunuel wrote:
The three digit positive integer N = a2b is both a multiple of 3 and a multiple of 5. What is N?

N=a2b
N is a multiple of 3 --> a + 2 + b = {a multiple of 3}.
N is a multiple of 5 --> b = 0 or b = 5.

If b = 0, then a can be 1, 4, or 7 (as a+2+b must be multiple of 3).
If b = 5, then a can be 2, 5, or 8 (for the same reason).

Total of 6 pairs of a and b thus 6 possible values of N.

(1) a + b is an even integer --> two pairs satisfy this b=0, a=4: N=420 OR b=5, a=5: N=525. Not sufficient.

(2) a/b = 1 --> only one possible combination for a and b: a=5, b=5 --> N=525. Sufficient.

Answer: B.


I dont understand how is that you have deduced that b must be a 0 or 5. I realize that any multiple of 5 has to have for its last digit either a 0 or 5. But here N = a2b is simply N = (a) (2) (b) and thus could have been written in a number of other ways. For example, N = 2ba.


N = a2b means that:
a = hundreds digit;
2 = tens digit;
b = units digit.

Hope it's clear.
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Re: The three digit positive integer N = a2b is both a multiple [#permalink] New post  18 May 2014, 10:10
MO B

N = a2b, where N is a multiple of 3 and a multiple of 5
so b must be 5 or 0


1)a+b is an even integer
when b=5, a=5
when b=0, a =4
not sufficient

2)a/b=1
in this case b not equal to 0. so b must be 5 and hence a must be 5
N is 525
sufficient
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Re: The three digit positive integer N = a2b is both a multiple [#permalink] New post  27 Dec 2021, 19:25
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Re: The three digit positive integer N = a2b is both a multiple [#permalink]
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