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Find the remainder when x = 5^5^5^5^5^5 ........24 times 5 is divided by 24 |
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Answer» since 5'5=25=odd no. 5'25=5*5*5*....=odd no. 5'oddno.=odd no. so x=5'5'5'5....24times =5*5*5*5*....oddtimes and since a no. 5'a can be written as 5*5*5*5......a times if a is to be odd no. then. 5'a/24=25*5*5*5...a-12times /24 =(1+1/24)*5*5*5*......a-2 times =5*5*.....+5*5*5.......a-2 times/24 =5*5*.....+25*5*5*5...a-4times/24 =5*5*...+(1+1/24)*5*5...a-4times. so if we keep doing this we will keep GETTING multiplication of fives a-2,a-4,a-6,.... 5,3,FINALLY 1 times so x/24= some integer +(1+1/24)*5 x/24 =integer+5/24 hence the ans is 5. the answer MAY seem a BIT of annoying but very simple arithmatic has been used. so if you find any difficulty you may ASK. |
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