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Difficulty:
95%
(hard)
Question Stats:
35% (03:10) correct
65%
(03:04)
wrong
based on 63
sessions
History
Date
Time
Result
Not Attempted Yet
How many possible pairs of values of (a,b) exist such that the number 49a81b is divisible by 66 ?
A. 1
B. 2
C. 3
D. 5
E. 7
Posted from my mobile device
A. 1
B. 2
C. 3
D. 5
E. 7
Posted from my mobile device
Updated on: 03 Mar 2020, 02:07
Originally posted by GMATinsight on 01 Mar 2020, 00:36.
Last edited by GMATinsight on 03 Mar 2020, 02:07, edited 1 time in total.
Last edited by GMATinsight on 03 Mar 2020, 02:07, edited 1 time in total.
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Divisiblity test of 6 - Number must be divisible by 2 as well as 3
Divisiblity test of 2 - Unit digit of the number must be even
Divisiblity test of 3 - Sum of the digits of the number must be divisible by 3
Divisibility test of 11- Difference of {Sum of even place digits} and {sum of odd place digits} of number must be 0 or divisible by 11
Number: 49a81b
i..e b must be even
4+9+a+8+1+b = 22+a+b must be divisible by 3
i.e. a+b must be either 2 or 5 or 8 or 11 or 14 or 17
Checking divisibility test of 11
Number: 49a81b
i.e. (4+a+1) - (9+8+b) must be 0 or divisible by 11
i.e. (17+b) - (5+a) = 12+(b-a) must be 0 or divisible by 11
i.e. b-a can be -1 only [P.S. a-b can have values only from -9 to 9]
since a-b is odd therefore a+b also should be odd and
since b must be even, therefore, a must be odd
FOr a+b = 5, (a,b) can be (3, 2) possible
FOr a+b = 17, (a,b) can be (9, 8) possible
i.e. there are two sets of values that are possible
Answer: Option B
02 Mar 2020, 10:05
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For a number to be divisible by composite number, we have to represent the composite number in its factorial form where each factors must be co-prime to each other.
Here for Number: 49a81b to be divisible by 66 , it must be divisible by 11,3 and 2.
Applying the rule of divisiblity of 2
b must be even.
Applying the rule of 3
4+9+a+8+1+b must be divisible by 3
Or , 22+ a+b must be divisible by 3. Hence possible values for a+b :
a+b = 2
a+b = 5
a+b = 8
a+b = 11
a+b = 14
a+b = 17
we should stop here because a+b =20 is not acceptable as a and b are single digit.
Now applying rule of 11
U-T = 0 , 11,22,33, .... etc
U= b+9+8
T = a+5
U-T = b+17-a-5
Or U-T = b-a+12
If we consider U-T = 0
Then a-b = 12 (not possible as a and b are single digit)
Now consider b-a+12=11
or, a-b = 1 (ok)
Now consider b-a+12=11
Or a-b =10(not possible as a and b are single digit)
Hence from rule of 11 we got the equation a-b =1
Now comparing the equations from rule of 3 and 11, we will reach in following scenarios:
a-b = 1
a+b = 2
Not possible as a cannot be 1.5
a-b = 1
a+b = 5
Solving we get a = 3 ; b= 2 ( 1st solution)
a+b = 8
a-b = 1
Not possible
a+b =11
a-b = 1
solving we get a= 6 ; b=5 . This is also not acceptable as b must be even (rule of 2 (66= 11X3X2))
a+b = 14
a- b = 1
Not possible
a+b = 17
a-b =1
Solving , we get a =9 ; b=8 ( 2nd solution)
Hence there are 2 possible solution for the pair (a,b)
(a,b) : (3,2) & (9,8)
Hence option B is correct.
This is a good question to clarify anyone's fundamental as multiple principles are applied here. I don't think this is a gmat oriented question but it's good for prep.
Here for Number: 49a81b to be divisible by 66 , it must be divisible by 11,3 and 2.
Applying the rule of divisiblity of 2
b must be even.
Applying the rule of 3
4+9+a+8+1+b must be divisible by 3
Or , 22+ a+b must be divisible by 3. Hence possible values for a+b :
a+b = 2
a+b = 5
a+b = 8
a+b = 11
a+b = 14
a+b = 17
we should stop here because a+b =20 is not acceptable as a and b are single digit.
Now applying rule of 11
U-T = 0 , 11,22,33, .... etc
U= b+9+8
T = a+5
U-T = b+17-a-5
Or U-T = b-a+12
If we consider U-T = 0
Then a-b = 12 (not possible as a and b are single digit)
Now consider b-a+12=11
or, a-b = 1 (ok)
Now consider b-a+12=11
Or a-b =10(not possible as a and b are single digit)
Hence from rule of 11 we got the equation a-b =1
Now comparing the equations from rule of 3 and 11, we will reach in following scenarios:
a-b = 1
a+b = 2
Not possible as a cannot be 1.5
a-b = 1
a+b = 5
Solving we get a = 3 ; b= 2 ( 1st solution)
a+b = 8
a-b = 1
Not possible
a+b =11
a-b = 1
solving we get a= 6 ; b=5 . This is also not acceptable as b must be even (rule of 2 (66= 11X3X2))
a+b = 14
a- b = 1
Not possible
a+b = 17
a-b =1
Solving , we get a =9 ; b=8 ( 2nd solution)
Hence there are 2 possible solution for the pair (a,b)
(a,b) : (3,2) & (9,8)
Hence option B is correct.
This is a good question to clarify anyone's fundamental as multiple principles are applied here. I don't think this is a gmat oriented question but it's good for prep.
