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08 Feb 2022, 05:51
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If the ten-digit integer \(a,390,484,58b\) is divisible by both 8 and 11, what is the value of ordered pair (a, b)?
(A) (8, 6)
(B) (7, 6)
(C) (6, 4)
(D) (5, 4)
(E) (4, 4)
Are You Up For the Challenge: 700 Level Questions
(A) (8, 6)
(B) (7, 6)
(C) (6, 4)
(D) (5, 4)
(E) (4, 4)
Are You Up For the Challenge: 700 Level Questions
What is the value of M and N respectively? If M39048458N is divisible by 8 and 11; Where M and N are single digit integers?
(1) 7, 8
(2) 8, 6
(3) 6, 4
(4) 5, 4
(1) 7, 8
(2) 8, 6
(3) 6, 4
(4) 5, 4
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kungfury42
Joined: 07 Jan 2022
Last visit: 31 May 2023
Posts: 580
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Schools: NUS '25 (A)
GMAT 1: 740 Q51 V38
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08 Feb 2022, 12:49
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Divisibility test by 8
Last 3 digits of the number must be divisible by 8
58b must be divisible by 8
560 is divisible by 8, add 24
584 is divisible by 8
Therefore, b=4
Divisibility test by 11
Let sum of odd placed digits = X
Let sum of even placed digits = Y
|X-Y| must be = 0 or 11n
X = 8+4+4+9+a = 25+a
Y = b+5+8+0+3 = 16+b = 20
|X-Y| = |25+a-20| = |5+a|
Since a can only be between [0, 9]
Only possible value of a is 6
Therefore, option C (6, 4)
Posted from my mobile device
Last 3 digits of the number must be divisible by 8
58b must be divisible by 8
560 is divisible by 8, add 24
584 is divisible by 8
Therefore, b=4
Divisibility test by 11
Let sum of odd placed digits = X
Let sum of even placed digits = Y
|X-Y| must be = 0 or 11n
X = 8+4+4+9+a = 25+a
Y = b+5+8+0+3 = 16+b = 20
|X-Y| = |25+a-20| = |5+a|
Since a can only be between [0, 9]
Only possible value of a is 6
Therefore, option C (6, 4)
Posted from my mobile device
19 Dec 2023, 02:41
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[quote="Bunuel"]If the ten-digit integer \(a,390,484,58b\) is divisible by both 8 and 11, what is the value of ordered pair (a, b)?
(A) (8, 6)
(B) (7, 6)
(C) (6, 4)
(D) (5, 4)
(E) (4, 4)
divisibility rule of 8 last 3 digits must be divisible by 8: hence 58b must be 584, b = 4
divisiblity rule of 11, even digits - odd digits = 0 / 11
hence a + 9 + 4 + 4 + 8 = 25, 3 + 0 + 8 + 5 + 4 = 20
a = 6 since 31 - 20 = 11 will make this divisible by 11
(A) (8, 6)
(B) (7, 6)
(C) (6, 4)
(D) (5, 4)
(E) (4, 4)
divisibility rule of 8 last 3 digits must be divisible by 8: hence 58b must be 584, b = 4
divisiblity rule of 11, even digits - odd digits = 0 / 11
hence a + 9 + 4 + 4 + 8 = 25, 3 + 0 + 8 + 5 + 4 = 20
a = 6 since 31 - 20 = 11 will make this divisible by 11
