(Q). Find the number of trailing zeros in (1! + 2! + 3! + 4! +5!)(6! + 7! + 8! + 9! + 10!)…………(96! + 97! + 98! + 99! + 100!).
Solution:-
(1! + 2! + 3! + 4! + 5!) ➝ as factorial number goes on increasing in every term.
Hence in every term number of trailing zeros will be the number of trailing zeros in the term’s lowest factorial number.
Hence
(1! + 2! + 3! + 4! + 5!) ➝ 0
(6! + 7! + 8! + 9! + 10!) ➝ 1
(11! + 12! + 13! + 14! + 15!) ➝ 2
(16! + 17! + 18! + 19! + 20!)➝ 3
(21! + 22! + 23! + 24! + 25!) ➝ 4
(26! + 27! + 28! + 29! + 30!) ➝ 6
(31! + 32! + 33! + 34! + 35!) ➝ 7
(36! + 37! + 38! + 39! + 40!) ➝ 8
(41! + 42! + 43! + 44! + 45!) ➝ 9
(46! + 47! + 48! + 49! + 50!)➝ 10
(51! + 52! + 53! + 54! + 55!) ➝12
(56! + 57! + 58! + 59! + 60!) ➝ 13
(61! + 62! + 63! + 64! + 65!) ➝ 14
(66! + 67! + 68! + 69! + 70!) ➝15
(71! + 72! + 73! + 74! + 75!) ➝ 16
(76! + 77! + 78! + 79! + 80!) ➝ 18
(81! + 82! + 83! + 84! + 85!) ➝ 19
(86! + 87! + 88! + 89! + 90!) ➝ 20
(91! + 92! + 93! + 94! + 95!) ➝ 21
(96! + 97! + 98! + 99! + 100!) ➝ 22
Now since expression is multiplication of these terms. Hence for value of expression all trailing zeros of individual terms will be added & this will be the final number of trailing zeros in the expression which is = 220 Answer