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08 Jul 2019, 08:23
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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:
41% (02:19) correct
59%
(02:19)
wrong
based on 153
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Which of the following cannot be the sum of a 6-digit number and the number formed by reversing its digits?
A. 777777
B. 1234442
C. 424523
D. 951467
E. 999999
A. 777777
B. 1234442
C. 424523
D. 951467
E. 999999
08 Jul 2019, 22:49
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shobhitkh
Let the numbers be abcdef and fedcba. When we add them we get
100000(a+f) + 10000(b+e) + 1000(c+d) + 100(d+c) +10(b+e) + (f+a)
= 100001(a+f) + 10010(b+e) + 1100(c+d)
Whatever (a+f) is, that will be the first digit and the last digit. (a+f) can be anything from 2 (=1+1) to 18 (=9+9). Option (A) works if (a+f) = 7, option (B) works if (a+f) = 12, option (C) works if (a+f) = 3 (and there is a carry over of 1 from b+e). Option (E) works if (a+f) = 9.
Option (D) does not work since if (a+f) = 7, the first digit cannot be 9. Even with a carry over, it can only be 8. Two single digits can never add up to give a carry over of 2.
Answer (D)
General Discussion
08 Jul 2019, 13:15
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Hi Bunuel VeritasKarishma
Can you please provide solution to the above problem?
Sum of the number and its reverse is a multiple of 11 but here all the numbers are divisible by 11. So I believe this concept doesn't apply here. How can we solve it?
Can you please provide solution to the above problem?
Sum of the number and its reverse is a multiple of 11 but here all the numbers are divisible by 11. So I believe this concept doesn't apply here. How can we solve it?
