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Q.The highest power of 2 by which the product of first 100 counting numbers can be divided without any remainder is_
a.97 b.96 c.95 d.94
(Please give the procedure for solving this ques)
first consider the perfect powers of 2..2,4,8,16,32,64their prod is divisible by=2^21nos remainig.. 6,12,14...100now the no. of prime nos below 50 which give numbers less that 100 on multiplying 2 leaving 2 are=15hence their prod is divisible by=2^15now the factors of 4 but not of any other perfect powers of 2 greater than 4,below 100, are12,20,28,36,.....92,100=AP with no. of terms=12so multiples of 2=2.12=24their prod is divisible by 2^24now consider multiples of 8 but not of any other perfect powers greater than 8they are=24,40,56,72,88=5 numbers=multiples of 2=5.3=15=divisible by 2^15now consider multiples of 16 but not of any other perfect powers of 2 greater than 16=48,80=2 numb.,so multiples of 2=2.4=8=2^8same for 32=96=2,for multiples of 2=1.5=2^5for 64 we have already considered it in perfect powers.hencethe total prod. is divisible by=2^(21+15+24+15+8+10)=2^(93)I THINK I MADE A LITTLE MISTAKE IN THE PROCESS.PLEASE COSS CHECK ONCEAND PLZ APPROVE.
the solution i mentioned was a conceptual approach,the formula for the highest power of a number x contained in n!is= [n!/x] + [n!/x^2] + [n!/x^3] + .....till it becomes 0.where [ ] denotes graetest integer function.here in this problem,n!=100!x=2so total power=[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64] =50+25+12+6+3+1=97 ANS
1*2*3..........*100 =100!
for power/exponent of 2 divide by 2,4,8,16,32.........
100/2=50 100/4=25 100/8=12 100/16=6 100/32=3 100/64=1
50+25+12+6+3+1=97
the formula for the highest power of a number x contained in n!is= [n!/x] + [n!/x^2] + [n!/x^3] + .....till it becomes 0.where [ ] denotes graetest integer function.here in this problem,n!=100!x=2so total power=[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64] =50+25+12+6+3+1=97 ANS
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