Elitmus
Exam
Numerical Ability
Number System
What s d max value of n if 224! Divisible by 224^n?
Read Solution (Total 8)
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- 224=2^5*7
224! How many sevens....because we have less sevens than 32......I.e 2^5in 224 factorial we have 36 times 7
So answer is 36...☑ - 8 years agoHelpfull: Yes(15) No(6)
- 224=(2^5)*7
consider for 7
224/7 =32
32/7 = 4
so n=32+4=36 - 8 years agoHelpfull: Yes(6) No(0)
- Prime factors of 224=2^5*7
Now,no. of 7s in 224 is given by
224/7 + 224/7^2 = 32+4 = 36
And the no. 2s in 224 is given by
224/2 + 224/2^2 + 224/2^3 + 224/2^4 +224/2^5 + 224/2^6 + 224/2^7 = 112 +56+28+14+7+3 = 220.
Therefore no. of 2^5 in 224 = 44.
Thus, n = 36.
(minimum of two is taken)
- 8 years agoHelpfull: Yes(5) No(1)
- answer will be 36
- 8 years agoHelpfull: Yes(1) No(2)
- 36 is the correct answer
- 8 years agoHelpfull: Yes(0) No(1)
- (A1+A2+A3+A4+A5+A6+7)=30*7=210
NOW AT TIME OF BIRTH OF YOUNGEST
(A1-7)+(A2-7)+(A3-7)+(A4-7)+(A5-7)+(A6-7)+(7-7)=210-49=161
NOW TOTAL 6 PERSON
{(A1-7)+(A2-7)+(A3-7)+(A4-7)+(A5-7)+(A6-7)}/6= 161/6 ans... - 8 years agoHelpfull: Yes(0) No(2)
- answer is : 36
224 = 2^5*7
224!/7 = 36 (so this is maximum power of 7 in 224!)
but if one power of 7 so power of 2 is 5
5*36=180 (power of 2 )
224!/2= pow>180
so clear ans is :36 - 7 years agoHelpfull: Yes(0) No(0)
- 224=(2^5)*7
In 224! , 2 comes 109 times
So (2^5) comes 20 times
And in 224! , 7 comes 36 times
Since 20 is less than 36 so the value of n is 20
Ans is 20 - 7 years agoHelpfull: Yes(0) No(1)
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