product of first ten natural numbers=2a×3b×5c×7d. then find the value of a+b+c+d
Answers
It has given that, product of first ten natural numbers = 2^a × 3^b × 5^c × 7^d
To find : The value of a + b + c + d
solution : first ten natural numbers are ; 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
product of first ten natural numbers = 2^a × 3^b × 5^c × 7^d
⇒1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 = 2^a × 3^b × 5^c × 7^d
⇒2 × 3 × 2² × 5 × (2 × 3) × 7 × 2³ × 3² × (2 × 5) = 2^a × 3^b × 5^c × 7^d
⇒2^8 × 3⁴ × 5² × 7¹ = 2^a × 3^b × 5^c × 7^d
on comparing we get,
a = 8 , b = 4 , c = 2 and d = 1
Therefore the value of (a + b + c + d) = 8 + 4 + 2 + 1 = 15
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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