Latest Elitmus Aptitude Question SOLUTION: What s d max value of n if 224! Divisible by 224^n?

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(#M40157067) ELITMUS Numerical Ability question elitmus quant 24august Keep an EYE

What s d max value of n if 224! Divisible by 224^n?

Asked In Elitmus Shivansh Shrivastava (7 years ago)

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Deepak singh  (7 years ago)

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...☑

Is this solution Helpfull? Yes (15) | No (6)  

Shashank Gupta  (7 years ago)

224=(2^5)*7

consider for 7
224/7 =32
32/7 = 4
so n=32+4=36

Is this solution Helpfull? Yes (6) | No   

Nidhi Mishra  (7 years ago)

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)

Is this solution Helpfull? Yes (5) | No (1)  

sachin pratap singh  (7 years ago)

answer will be 36

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Vikash kumar  (7 years ago)

36 is the correct answer

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lekhraj jeengar  (7 years ago)

(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...

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sandeep  (7 years ago)

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

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megha jain  (6 years ago)

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

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