Solution of the Aptitude question- n o o n s o o n m o o n --------------- j u n e --------------- step1: when you add n, n ,n you got e so e must me a multiple of 3 from 1 to 9, multiple of 3 are 3,6,9 if e = 3, n = 1, if e = 6, n = 2, if e = 9, n = 3. step2: when u add o,o,o you got n. tricky part is here n is just the units place, if o+o+o is 24 then n = 4 and 2 is send as carry... i know u know additions, else take madhus help.:) assuming e = 3, n = 1 , when u add o,o,o you must again get multiple of 3, but here they can be greater than 9, but less than 27, since, o can not be greater than 9. so possible multiples are 3,6,9,12,15,18,21,24,27. since we are assuming e = 3, n = 1, to get 1 in ns place of june, only possible multiple of 3 is 21. satisfies, n = 1 and 2 is send as carry. to get 21, o must be 7, bcaz 7+7+7 = 21 so now we got, n = 1, e = 3, o = 7. step3: we got so far, n = 1, e = 3, o = 7. here again, we add, o+o+o+2(carry from 21) and we get u and carry is forwarded. since o = 7, 21+2 = 23, which makes u = 3 and 2 as carry, buttttttttt, already we have e = 3 ,so u can not be equal to 3, so our assumption in step 2, e = 3, n = 1 is false not we have to try with remaining two assumptions.... dont sleep...... look at the remaining part tooo. other two assumptions are if e = 6, n = 2, if e = 9, n = 3. now step 4: assuming if e = 6, n = 2, when u add o,o,o you must again get multiple of 3, but here they can be greater than 9, but less than 27, since, o can not be greater than 9. so possible multiples are 3,6,9,12,15,18,21,24,27. since we are assuming e = 6, n = 2,, to get 2 in ns place of june, only possible multiple of 3 is 12. satisfies, n = 2 and 1 is send as carry. to get 12, o must be 4, bcaz 4+4+4 = 12, o = 4, n = 2, e = 6 step5: so far, we got o = 4, n = 2, e = 6 here again, we add, o+o+o+1(carry from 12) and we get u and the new carry is forwarded. since o = 4, 12+1 = 13, which makes u = 3 and 1 as carry, so, u =3, o = 4, n = 2, e = 6 all are unique so no problem... we can go further and you dont yawnnn step 6: till now we got u =3, o = 4, n = 2, e = 6 here, n+s+m+1(carry from 13) is j so, 2+s+m+1 = j from1 to 9, the reamining numbers are 1,5,7,8,9. since 2,3,4,6 are used by n,u,o,e respectively here, 3+s+m = j, so j is greater than 3, so the possible numbers for j are now, 5,7,8,9. assume j = 5, 3+s+m =5, which means, s = m =1, uniquness is not achieved, so j != 5, assume j = 7, 3+s+m = 7, means s+m =4 here, possbile values of s and m to satisfy s+m =4, are 1+3 =4 not possible since u = 3 2+2 = 4 not possible since both cant be equal 3+1 =4 not possible since u = 3 assume j = 8 3+s+m = 8, means s+m = 5 possbile values of s and m to satisfy s+m =5, are 1+4 = 5 not possible since o = 4 2+3= 5 not possible since u = 3 3+2 = 5 not possible since u = 3 4+1 + 5 not possible since o = 4 therefore j = 9 3+s+m = 9 s+m = 6 s = 1 and m =5 or s = 5 and m = 1 n o o n 2 4 4 2 s o o n 1 4 4 2 m o o n -------------- 5 4 4 2 --------------- -------------- ---------------- j u n e 9 3 2 6 --------------- ---
Posted on: Thu, 19 Jun 2014 17:59:11 +0000
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