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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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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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\(a^2\) 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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