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bhanu29
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If a and b are consecutive positive integers such that a2 is divisible Updated on: 31 Oct 2024, 08:28
Originally posted by bhanu29 on 30 Oct 2024, 20:25.
Last edited by bhanu29 on 31 Oct 2024, 08:28, edited 1 time in total.
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C
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38% (02:26) correct 62% (02:21) wrong based on 134 sessions

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If \(a\) and \(b\) are consecutive positive integers such that \(a^2\) is divisible by \(25\) and \(b^2\) is divisible by \(63\), which of the following is the greatest integer that must be a factor of \(ab\) ?

(A) 60
(B) 105
(C) 210
(D) 225
(E) 315

Edit: Updated Correct answer, sorry for the confusion.
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C
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kingbucky
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If a and b are consecutive positive integers such that a2 is divisible 30 Oct 2024, 21:58
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\(a^2\) is divisible by 25
implies \(5\) divides \(a\)
\(b^2\) is divisible by 63
implies \(3\) divides \(b\), \(7\) divides \(b\)
\(a\) and \(b\) are consecutive, that means either of them is even ie \(2\) divides \(ab\)
therefore \(5*3*7*2\) divides \(ab\)
ie \(210\) divides \(ab\).

checking by taking \(a =20, b=21\),
we have \(210\) divides \(ab\), whereas options \(225, 315\) do not divide. therefore (c) should be the correct answer.
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bhanu29
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If a and b are consecutive positive integers such that a2 is divisible 31 Oct 2024, 08:29
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The answer is \(210\) (updated the answer)
raj-boro
the answer is 225 indeed.
If a^2/25, it gives us 5k.
If b^2/63, it gives us 63k.
For a and b to be consecutive----->
5*25=125=a
63*2=126=b
a,b=125,126
Now,
125=5^3 and
126=2*3^2*7
a*b= 5^3 *2*3^2*7
Since the question is to find the greatest integer, let factorize from the last options-

315=3^2*5*7.
Now dividing a*b= 5^3 *2*3^2*7 by 315=3^2*5*7 leaves us with 50.
225= 3^2*5^2
Now dividing a*b= 5^3 *2*3^2*7 by 225= 3^2*5^2 leaves us with 70.

Therefore, 225 is the greatest integer that must be a factor of a*b.

Hope you find this helpful.
Happy learning. :)
bhanu29
If \(a\) and \(b\) are consecutive positive integers such that \(a^2\) is divisible by \(25\) and \(b^2\) is divisible by \(63\), which of the following is the greatest integer that must be a factor of \(ab\) ?

(A) 60
(B) 105
(C) 210
(D) 225
(E) 315
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If a and b are consecutive positive integers such that a2 is divisible 31 Oct 2024, 08:39
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a^2 is divisible by 25 (5*5). Hence, a must be a factor of 5.
b^2 is divisible by 63 (3*3*7). Hence, b must be a factor of 21.
For a & b to be consecutive integers, so one must be even & one must be odd.
5*21*2=210

Answer: C
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If a and b are consecutive positive integers such that a2 is divisible 31 Oct 2024, 09:02
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yes, seems i got wrong there.
Consider not minding my reply. :)
bhanu29
The answer is \(210\) (updated the answer)
raj-boro
the answer is 225 indeed.
If a^2/25, it gives us 5k.
If b^2/63, it gives us 63k.
For a and b to be consecutive----->
5*25=125=a
63*2=126=b
a,b=125,126
Now,
125=5^3 and
126=2*3^2*7
a*b= 5^3 *2*3^2*7
Since the question is to find the greatest integer, let factorize from the last options-

315=3^2*5*7.
Now dividing a*b= 5^3 *2*3^2*7 by 315=3^2*5*7 leaves us with 50.
225= 3^2*5^2
Now dividing a*b= 5^3 *2*3^2*7 by 225= 3^2*5^2 leaves us with 70.

Therefore, 225 is the greatest integer that must be a factor of a*b.

Hope you find this helpful.
Happy learning. :)
bhanu29
If \(a\) and \(b\) are consecutive positive integers such that \(a^2\) is divisible by \(25\) and \(b^2\) is divisible by \(63\), which of the following is the greatest integer that must be a factor of \(ab\) ?

(A) 60
(B) 105
(C) 210
(D) 225
(E) 315
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If a and b are consecutive positive integers such that a2 is divisible 01 Apr 2025, 10:52
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\(a^2\) is divisible by \(25\), it means a is divisible by 5

\(b^2\) is divisible by \(63\), it means b is a factor of 7*3 = 21

Given a and b are consecutive, a is at least 20

ab = 20*21 = 420
A, B and C divides 420 but not D and E. Of A,B and C, C is greatest

bhanu29
If \(a\) and \(b\) are consecutive positive integers such that \(a^2\) is divisible by \(25\) and \(b^2\) is divisible by \(63\), which of the following is the greatest integer that must be a factor of \(ab\) ?

(A) 60
(B) 105
(C) 210
(D) 225
(E) 315

Edit: Updated Correct answer, sorry for the confusion.
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