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12 May 2019, 23:43
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Difficulty:
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Question Stats:
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How many integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999? For example, both 121 and 211 have this property.
A. 226
B. 243
C. 270
D. 469
E. 486
A. 226
B. 243
C. 270
D. 469
E. 486
22 May 2019, 04:36
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Permutation of its digits is a multiple of 11.
Eg: 121 is a multiple so the set includes 112,121,211
Multiples of 11 between 100 and 999:
110 is the 990 is the least and greatest multiple in the range
No. of multiples = (990-110)/11 +1 = 81
So the maximum number of integers the set can have = 81*3 = 243
Eliminate C,D,E
For multiples of 10, only two permutations are possible
Eg: 990 --> 909,990
So, subtract 9 from 243 = 234
Eliminate B
Ans is A
Posted from my mobile device
Eg: 121 is a multiple so the set includes 112,121,211
Multiples of 11 between 100 and 999:
110 is the 990 is the least and greatest multiple in the range
No. of multiples = (990-110)/11 +1 = 81
So the maximum number of integers the set can have = 81*3 = 243
Eliminate C,D,E
For multiples of 10, only two permutations are possible
Eg: 990 --> 909,990
So, subtract 9 from 243 = 234
Eliminate B
Ans is A
Posted from my mobile device
22 May 2019, 12:48
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All three digit multiples of 11 can be written in three different ways except multiples that contain digit 0.
Multiples of 11 with digit 0 can be written in two different ways. eg. 110-101-011. Now 011 is not included between 100 and 999.
100-199:
Multiples: 110, 121, 132, 143, 154, 165, 176, 187, 198.
Note that there is only one multiple with 0.
No of rotations: 8*3 + 1*2= 26.
Now, from 200 to 999, every other group of 100 integers will contain two such multiples containing 0. Eg 209, 220.
No. of rotations per 100 integers in these groups= 7*3 + 2*2 = 25
Total groups: 8
No. of rotations in 800 integers from 200 to 999: 25*8= 200
Total rotations in 900 integers= 26+200= 226
Ans A
Posted from my mobile device
Multiples of 11 with digit 0 can be written in two different ways. eg. 110-101-011. Now 011 is not included between 100 and 999.
100-199:
Multiples: 110, 121, 132, 143, 154, 165, 176, 187, 198.
Note that there is only one multiple with 0.
No of rotations: 8*3 + 1*2= 26.
Now, from 200 to 999, every other group of 100 integers will contain two such multiples containing 0. Eg 209, 220.
No. of rotations per 100 integers in these groups= 7*3 + 2*2 = 25
Total groups: 8
No. of rotations in 800 integers from 200 to 999: 25*8= 200
Total rotations in 900 integers= 26+200= 226
Ans A
Posted from my mobile device
