Find the value of a+b+c+d if the product frist 10 natural numbers is written as 2a×3b×5c×7d
Answers
Step-by-step explanation:
product of first ten natural numbers=1x2x3x4x5x6x7x8x9x10
=1x2x3x2x2x5x2x3x7x2x2x2x3x3x2x5
=2^8. x. 3^4. x. 5^2. x. 7^1
=2x2^7. x. 3x3^3. x. 5x5. x. 7x1=2ax3bx5cx7d
by comparing on both sides we get,
a=2^7=128
b=3^3=27
c=5
d=1
thus,a+b+c+d=128+27+5+1=161
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