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29 Jul 2022, 02:32
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E
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Difficulty:
75%
(hard)
Question Stats:
60% (02:50) correct
40%
(02:46)
wrong
based on 129
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A five-digit integer 7a1b0, where a and b are thousands and tens digits, respectively, is divisible by 72. If |a - b| is a prime number, what is the value of a*b ?
A. 8
B. 12
C. 16
D. 18
E. 24
A. 8
B. 12
C. 16
D. 18
E. 24
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29 Jul 2022, 12:56
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Bunuel
\(7a1b0\) is divisible by \(72\).
\(72 = 2^3 * 3^2\) => \(72 = 8 * 9\)
So, \(7a1b0\) is divisible by both \(8\) and \(9\)
Divisibility rule for \(8\): Last 3 digits need to be divisible by 8
Divisibility rule for \(9\): Sum of digits need to be divisible by 9
For \(7a1b0\) to be divisible by 8, \(b\) needs to be either \(2\) or \(6\) because only \(1\) and \(0\) are fixed and only \(120\) and \(160\) are divisible by 8
So: If \(b = 2\), \(|a - b| = |a - 2| is prime => a = 4/5/7/9\)
In any of the above cases for \(a\), sum of digits will not be divisible by 9 (\(14/15/17/19\))
So: If \(b = 6\), \(|a - b| = |a - 6| is prime => a = 1/3/4/8/9\)
In all the above cases, only for \(a = 4\) is the sum of digits divisible by 9
So, overall, we only have 1 case in which the number is divisible by 72 => \(a = 4, b = 6\)
Product = \(ab = 24\)
Answer - E
14 Aug 2022, 09:35
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Sol: Sum of Digit = 8+a+b ( should be multiple/divisible of 9 as 72=8x9) >> a+b = 1 (1,0), 10 {(1,9) (2,8),(3,7), (4,6), (5,5)} 19...
a+b can not equal to 1 since |a-b| = prime
now using {(1,9) (2,8),(3,7), (4,6), (5,5)} in |a-b|= prime
{(1,9) = 8 no
(2,8)= 6 no,
(3,7)= 4 no
(4,6) = 2 yes,
(5,5) = 0 no
axb= 4x6= 24
E ans
a+b can not equal to 1 since |a-b| = prime
now using {(1,9) (2,8),(3,7), (4,6), (5,5)} in |a-b|= prime
{(1,9) = 8 no
(2,8)= 6 no,
(3,7)= 4 no
(4,6) = 2 yes,
(5,5) = 0 no
axb= 4x6= 24
E ans
