Can anyone help me please
if we write 88! in base 8, how many zeros will be in the end of the number ?
if we write 88! in base 8, how many zeros will be in the end of the number ?
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I love these math-team type questions! First, you need to recognise what a zero means at the end of a base-8 number. Perhaps pattern recognition will help:
10 in base 8 is equal to 8^1. 20 is just twice that, so it is still divisible by 8^1.
100 in base 8 is equal to 8^2. 200 is just twice that, so it is still divisible by 8^2.
1000 in base 8 is equal to 8^3. 2000 is just twice that, so it is still divsiible by 8^3.
So, it seems that if there are n zeros at the end of a number written in base 8, the number is divisible by 8^n. You can prove this if you like, but it seems fairly reasonable. In fact, if there are n zeros at the end of a number written in base k, then the number is divisible by k^n. (Think of base k=10...it makes PERFECT sense now!)
So, we want to figure out the HIGHEST power of 8 that divides 88!. Or...how about we find out the highest power of 2 instead, and then divide that by 3 (discarding the leftovers)? The reason I want to do that will be clear later. But let's say that 2^10 divides a number but 2^11 doesn't. Then that means that there are 10/3 = 2.333 powers of 8, or 2 powers of 8, that divide the number.
So, the number is 1*2*3*4*...*88.
The numbers 2, 4, 6, 8, ..., 88 each donate at least one power of 2. That's 44 powers of 2.
The numbers 4, 8, 12, ..., 88 each donate at least one MORE power of 2. That's 22 powers of 2.
The numbers 8, 16, 24, 32...88 each donate at least one MORE power of 2. That's 11 powers of 2.
The numbers 16, 32, 48, 64, and 80 each donate one more power of 2. That's 5 powers of 2.
The numbers 32 and 64 each donate one more power of 2. That's 2 more powers of 2.
And finally, 64 donates one more power of 2. That's 1 power of 2.
So your answer is 44 + 22 + 11 + 5 + 2 + 1 powers of 2 are in 88!. Divide that by 3, and then round down, because you're looking for powers of 8, not powers of 2.
10 in base 8 is equal to 8^1. 20 is just twice that, so it is still divisible by 8^1.
100 in base 8 is equal to 8^2. 200 is just twice that, so it is still divisible by 8^2.
1000 in base 8 is equal to 8^3. 2000 is just twice that, so it is still divsiible by 8^3.
So, it seems that if there are n zeros at the end of a number written in base 8, the number is divisible by 8^n. You can prove this if you like, but it seems fairly reasonable. In fact, if there are n zeros at the end of a number written in base k, then the number is divisible by k^n. (Think of base k=10...it makes PERFECT sense now!)
So, we want to figure out the HIGHEST power of 8 that divides 88!. Or...how about we find out the highest power of 2 instead, and then divide that by 3 (discarding the leftovers)? The reason I want to do that will be clear later. But let's say that 2^10 divides a number but 2^11 doesn't. Then that means that there are 10/3 = 2.333 powers of 8, or 2 powers of 8, that divide the number.
So, the number is 1*2*3*4*...*88.
The numbers 2, 4, 6, 8, ..., 88 each donate at least one power of 2. That's 44 powers of 2.
The numbers 4, 8, 12, ..., 88 each donate at least one MORE power of 2. That's 22 powers of 2.
The numbers 8, 16, 24, 32...88 each donate at least one MORE power of 2. That's 11 powers of 2.
The numbers 16, 32, 48, 64, and 80 each donate one more power of 2. That's 5 powers of 2.
The numbers 32 and 64 each donate one more power of 2. That's 2 more powers of 2.
And finally, 64 donates one more power of 2. That's 1 power of 2.
So your answer is 44 + 22 + 11 + 5 + 2 + 1 powers of 2 are in 88!. Divide that by 3, and then round down, because you're looking for powers of 8, not powers of 2.
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