The number 4,000,000 has 63 positive integral factors. How do you find a and b, where 2^a 5^b is the product of all positive factors of 4,000,000?

Answer 1

#a=8# and #b=6#

#4000000=4xx10xx10xx10xx10xx10xx10#
= #2^2xx10^6#
= #2^2xx(2xx5)^6#
= #2^2xx2^6xx5^6#
= #2^(2+6)xx5^6#
= #2^8xx5^6#
Comparing it with #2^a5^b#, we get
#a=8# and #b=6#
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Answer 2

#a=252# and #b=189#

#4000000 = 4 * 10^6 = 2^8*5^6#
So each of the positive integral factors of #4000000# is of the form #2^m*5^n# where #m in { 0, 1, 2, 3, 4, 5, 6, 7, 8 }# and #n in {0, 1, 2, 3, 4, 5, 6}#.

Take note of this:

#sum_(m=0)^8 m = 1/2 8 (8+1) = 36#
#sum_(n=0)^6 n = 1/2 6 (6+1) = 21#

Below, we make use of these...

The sum of the individual factors is:

#prod_(m=0)^8 (prod_(n=0)^6 2^m*5^n)#
#= prod_(m=0)^8 (2^(7m) prod_(n=0)^6 5^n)#
#= (prod_(m=0)^8 2^(7m)) (prod_(n=0)^6 5^n)^9#
#= 2^(7sum_(m=0)^8 m)*5^(9 sum_(n=0)^6 n#
#= 2^(7*36)*5^(9*21)#
#= 2^252 * 5^189#
So #a=252# and #b=189#
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Answer 3

To find ( a ) and ( b ), where ( 2^a \times 5^b ) is the product of all positive factors of ( 4,000,000 ), first, you need to express ( 4,000,000 ) in terms of its prime factors.

( 4,000,000 = 2^6 \times 5^6 )

Next, to determine the exponents ( a ) and ( b ), observe that each factor can either include a ( 2 ) or a ( 5 ), or both. Since ( 4,000,000 ) has ( 63 ) positive integral factors, and it can be represented as ( 2^6 \times 5^6 ), each factor could have any combination of the prime factors ( 2 ) and ( 5 ) with exponents ranging from ( 0 ) to ( 6 ).

Given that ( 63 ) is the total number of factors, there are ( 7 ) choices for the exponent of ( 2 ) (( 0 ) through ( 6 )) and ( 7 ) choices for the exponent of ( 5 ) (( 0 ) through ( 6 )).

Therefore, the total number of combinations of ( a ) and ( b ) is ( 7 \times 7 = 49 ).

As ( 2^6 \times 5^6 ) is included in this count, subtract ( 1 ) to get the actual number of combinations of ( a ) and ( b ).

Thus, ( a ) can be any integer from ( 0 ) to ( 6 ), and ( b ) can also be any integer from ( 0 ) to ( 6 ).

Therefore, ( a ) can take on ( 7 ) different values, and ( b ) can also take on ( 7 ) different values, giving a total of ( 7 \times 7 = 49 ) combinations.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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