Question

Find IGP of 8 in 120!

Answer 1

Question :- Find IGP(Highest power) of 8 in 120! ?

Solution :-

as we know that, 8 can be written as 2³ .

now, we know that,

• If p is a prime number, then highest power of p^a present in a factorial n is given by = (Highest power of p in n!) / a .

we have given that,

• p = 2
• n = 120
• a = 3 .

so, Highest power of 2³ in 120! :-

→ (Highest power of 2 in 120!) / 3

→ [(120/2¹) + (120/2²) + (120/2³) + (120/2⁴) + (120/2⁵) + (120/2⁶) + (120/2⁷)] / 3

→ [(120/2) + (120/4) + (120/8) + (120/16) + (120/32) + (120/64) + (120/128)] / 3

→ [ 60 + 30 + 15 + 7 + 3 + 1 + 0 ] / 3

→ (116/3)

→ 38 . (Ans.)

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Answer 2

Answer: 38

Step-by-step explanation: