Question

find the value of a+b+c+d ,if product of first ten natural. numbers is written as 2ax3bx5cx7d​

Answer 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