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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)

8 years ago

first consider the perfect powers of 2..
2,4,8,16,32,64
their prod is divisible by=2^21
nos remainig.. 6,12,14...100
now the no. of prime nos below 50 which give numbers less that 100 on multiplying 2 leaving 2 are=15
hence their prod is divisible by=2^15
now the factors of 4 but not of any other perfect powers of 2 greater than 4,below 100, are
12,20,28,36,.....92,100=AP with no. of terms=12
so multiples of 2=2.12=24
their prod is divisible by 2^24
now consider multiples of 8 but not of any other perfect powers greater than 8
they are=24,40,56,72,88=5 numbers=multiples of 2=5.3=15=divisible by 2^15
now 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=2^8
same for 32=96=2,for multiples of 2=1.5=2^5
for 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.
AND PLZ APPROVE.

8 years ago

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=2
so total power=[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64]
=50+25+12+6+3+1=97 ANS

8 years ago

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

8 years ago

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=2
so total power=[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64]
=50+25+12+6+3+1=97 ANS

3 years ago
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=2
so total power=[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64]
=50+25+12+6+3+1=97 ANS