Math, asked by praveenjaydarling530, 8 months ago

product of first ten natural numbers =2a×3b×5c×7d then find the value of a+b+c+d​

Answers

Answered by rahulchaudhary14
10

Step-by-step explanation:

first ten natural number=1,2,3,4,5,6,7,8,9,10

1×2×3×4×5×6×7×8×9×10=2a×2b×2c×2d

3628800=2(abcd)

abcd=1814400 answer

Answered by abhi569
1

Answer:

161  

Step-by-step explanation:

First ten natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Product(even no.) = 2*4*6*8*10

                              = 2⁵(1*2*3*4*5)

                              = 2⁸ * 3¹ * 5¹

Product(odd no.) = 1*3*5*7*9

                            = 3³ * 5 * 7

Product of all = product of even * odd

              = 2⁸ * 3¹ * 5¹ * 3³ * 5 * 7

              = 2⁸ * 3⁴ * 5² * 7

Compare this with 2a*3b*5c*7c:

2a = 2⁸    &    3b = 3⁴   &  

5c = 5²    &  7d = 7

     Thus,

a = 2⁸/2 = 2⁷ = 128  

b = 3³=27  ,  c = 5    and d = 1

 Hence, a + b + c + d = 128+27+5+1

                                 = 161

If it is 2^a x 3^b x 5^c x 7^d​, answer is 8+4+2+1=15

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