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Q. Let N be a 5 digit no. that reverses itself on multiplication with 4 that is N=abcde & abcde*4 = edcba. a,b,c,d,e are distinct digits.then what is the sum of digits of N?

 

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(a) 30 (b) 18 (c)27 (d)36

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abcde*4=edcba

a= 1 or 2 if >2 will give a carry over and it will result in 6 digit number
now ba should be divisible by 4 so a=2 (cant be 1)

now ba can be 12,32… so b =1 or 3
also e has to end with 2 at its units place …so it can be 3 or 8 (3*4=12 and 4*8=32)

if ba=12 we get e=8 therefore b=1

now d8*4 should be 12 so d can so when 4*8=32 carry over 3 now d must be 7 as 4*7=28+3 = 31

21X78*4=87X12

so X=3,5,6,9
so trial and error methods se X=9

so answer is 21978*4=87912

Sum of digits = 27

Answer given by Vaibhav Autade

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