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A!*B!*C!=D!, A,B,C are all odd natural numbers and D=10,what is A*B*C?

189

105

315

27

Please share approach too,

189

105

315

27

Please share approach too,

0 votes

Best answer

By reading the question we can assume that A, B, C have to be less than 10 and they all are ODD.

AND

**10! = 2 ^{8} * 3^{4} * 5^{2} * 7^{1}**

Now, we'll be factorising the factorials of Odd natural numbers which are less than 10.

1! = 2^{0} * 3^{0} * 5^{0} * 7^{0}

3! = 2^{1} * 3^{1} * 5^{0} * 7^{0}

5! = 2^{3} * 3^{1} * 5^{1 }* 7^{0}

7! = 2^{4} * 3^{2} * 5^{1} * 7^{1}

9! = 2^{7} * 3^{4} * 5^{1} * 7^{1}

Now, we have to identify three powers of 2 from the above one so that they all add up to 8,.

and we're getting 2 possibilities

**1.** (0, 1, 7)

or

**2.** (1, 3, 4)

Considering the 1st one:

(0, 1, 7)

**∴ **A will be 1

B will be 3

C will be 9

And multiplying A!*B!*C! = 1! * 3! * 9! = 2^{8 }* 3^{5 }* 5^{1} * 7^{1 }= 2177280 (we can ignore this big multiplication and can see the factors are not matching with 10! factors).

So, we can't accept this case.

Now, the 2nd case

(1, 3, 4)

**∴ **A will be 3

B will be 5

C will be 7

A!*B!*C! = 3! * 5! * 7! = 2^{8} * 3^{4} * 5^{2} * 7^{1} which is equal to 10!

**A*B*C = 3 * 5 * 7 = 105**

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