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04 Nov 2021, 05:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (02:34) correct
69%
(02:54)
wrong
based on 71
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How many 5 digit numbers can be formed using the digits 1, 2, 3, 4, 5 and 6 without repetition that are divisible by 8?
(A) 56
(B) 64
(C) 72
(D) 84
(E) 96
(A) 56
(B) 64
(C) 72
(D) 84
(E) 96
04 Nov 2021, 05:10
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numbers are divisible by 8 when their last 2 digits are divisible by 8, so the scenarios are:
- - - - 24
- - - - 32
- - - - 16
- - - - 64
now for each case we have to consider 4! situations, thus 4*(4!) = 96
the answer is E, IMO
- - - - 24
- - - - 32
- - - - 16
- - - - 64
now for each case we have to consider 4! situations, thus 4*(4!) = 96
the answer is E, IMO
04 Nov 2021, 06:03
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First, if you had to guess here, we can make 6*5*4*3*2 = 720 numbers in total, and roughly 1/8 of them should be divisible by 8, so the answer should be close to 90 (that estimate is only good here because our digits are half-even, half-odd, so we wouldn't expect our numbers to be predominantly divisible by 8 or not). So D and E are the best guesses, but the answer can be slightly different than 90 because of the precise digits we're given.
For a number to be divisible by 8, its last three digits must form a number divisible by 8. So for example, since 432 is divisible by 8, the number 56,432 and 61,432 will both be divisible by 8. The question now becomes really tedious, because we need to identify all of the three-digit combinations we can make, using the digits provided, that will be divisible by 8. If our number is divisible by 8, it must be even, so the last digit will be even. And our number must be divisible by 4, so the last two digits will need to form a multiple of 4, and our last two digits must be 12, 32, 52, 24, 64, 16, 36 or 56. Finally to work out what 3-digit numbers we can now make that will be multiples of 8, it can be helpful to notice: if you start with a multiple of 8, say '24', and add 100, you never get a multiple of 8, and if you add 200 you always do (because 100 is not divisible by 8, but 200 is), and if you instead start with a multiple of 4 that is not a multiple of 8, then adding 100 always gives you a multiple of 8 but adding 200 does not. Using that observation we can, beginning from our two-digit multiples of 4, work out all the three-digit numbers we can make that are multiples of 8:
12 --> 312, 512
32 --> 432, 632
52 --> 152, 352
24 --> 624
64 --> 264
16 --> 216, 416
36 --> 136, 536
56 --> 256, 456
Remember we can't repeat digits, so we shouldn't count numbers like 224 or 112 at this point. So we have 14 numbers we can make that are divisible by 8, and once we make any of those 14 numbers, we'll have 3 remaining digits to choose from for the ten thousands digit, and then 2 remaining choices for the thousands digit, and the answer is 14*3*2 = 84.
I did this a bit quickly and when itemizing all these cases it's easy to make a careless error and miss a possibility, so hopefully I haven't done that, but this is not the kind of thing you need to do on the GMAT.
For a number to be divisible by 8, its last three digits must form a number divisible by 8. So for example, since 432 is divisible by 8, the number 56,432 and 61,432 will both be divisible by 8. The question now becomes really tedious, because we need to identify all of the three-digit combinations we can make, using the digits provided, that will be divisible by 8. If our number is divisible by 8, it must be even, so the last digit will be even. And our number must be divisible by 4, so the last two digits will need to form a multiple of 4, and our last two digits must be 12, 32, 52, 24, 64, 16, 36 or 56. Finally to work out what 3-digit numbers we can now make that will be multiples of 8, it can be helpful to notice: if you start with a multiple of 8, say '24', and add 100, you never get a multiple of 8, and if you add 200 you always do (because 100 is not divisible by 8, but 200 is), and if you instead start with a multiple of 4 that is not a multiple of 8, then adding 100 always gives you a multiple of 8 but adding 200 does not. Using that observation we can, beginning from our two-digit multiples of 4, work out all the three-digit numbers we can make that are multiples of 8:
12 --> 312, 512
32 --> 432, 632
52 --> 152, 352
24 --> 624
64 --> 264
16 --> 216, 416
36 --> 136, 536
56 --> 256, 456
Remember we can't repeat digits, so we shouldn't count numbers like 224 or 112 at this point. So we have 14 numbers we can make that are divisible by 8, and once we make any of those 14 numbers, we'll have 3 remaining digits to choose from for the ten thousands digit, and then 2 remaining choices for the thousands digit, and the answer is 14*3*2 = 84.
I did this a bit quickly and when itemizing all these cases it's easy to make a careless error and miss a possibility, so hopefully I haven't done that, but this is not the kind of thing you need to do on the GMAT.
