product of first ten natural numbers=2a×3b×5c×7d. then find the value of a+b+c+d
Answers
Question :- product of first ten natural numbers = 2^a×3^b×5^c×7^d. then find the value of (a+b+c+d). ?
Solution :-
As we know that, natural numbers starts from 1.
So,
First 10 natural numbers are upto 10 .
A/q,
→ 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 = 2^a * 3^b * 5^c * 7^d
→ 1 * 2 * 3 * (2 * 2) * 5 * (2 * 3) * 7 * (2 * 2 * 2) * (3 * 3) * (2 * 5) = 2^a * 3^b * 5^c * 7^d
→ 1 * (2 * 2 * 2 * 2 * 2 * 2 * 2 * 2) * (3 * 3 * 3 * 3) * (5 * 5) * 7 = 2^a * 3^b * 5^c * 7^d
using {a^m * a^n = a^(m + n)} ,
→ 1 * (2^8) * (3⁴) * (5²) * 7 = 2^a * 3^b * 5^c * 7^d
→ (2^8) * (3⁴) * (5²) * 7 = 2^a * 3^b * 5^c * 7^d
comparing Now, we get :-
- a = 8
- b = 4
- c = 2
- d = 1
Therefore,
→ a + b + c + d
→ 8 + 4 + 2 + 1
→ 15 (Ans.)
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