MBA
Exam
Que: N=202*20002*200000002*20000000000000002*20000......2 (31 zeros).The sum of digits in this multiplication will be: 1) 112 2) 144 3) 160 4) None
Read Solution (Total 1)
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202 = 202 (there are two 2's)
202 * 20002 = 4040404 (there are four 4's)
So, if you see, there are four 4's in the above multiplication, i.e., (number of zeroes + 1 ) number of fours.
Similarly, for 202 * 20002 * 200000002 = there will be eight 8's.
For 202 * 20002 * 200000002 * 20000 ...2 (15 zeroes) = there will be sixteen 16's.
For 202 * 20002 * 200000002 * 20000 ...2 (15 zeroes) * 2000... 2 (31 zeros) = there will be thirty two 32's.
Thirty two 32's contains thrity two 3's and thirty two 2's.
So, 32 * 3 + 32 * 2 = 160 :)
- 9 years agoHelpfull: Yes(0) No(0)
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