What is the smallest number with which 200 should be multiplied to make it divisible by 12100?

Question 2 Exercise 3.5

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Answer:

SOLUTION:

The smallest four prime numbers are 2,3,5 and 7.

hence, the required number is 2\times3\times5\times7=210

Video transcript

"Hello guys, welcome to Lido homework today we'll do question number 12. Now. This is when I questioned or five immigration. So the question is I am the smallest prime number having four different prime factors. Can you find me? So as the numbers for different prime factors? So take the for smallest conjugate of prime numbers. So the for smallest conjugate of prime numbers are 2 3 5 and 7. So these are the prime factors of that of this number that we've defined. So as these are prime factors when you multiply all of these As you're going to get that number, so your answer is 2 into 3. Into 5 into 7 Therefore, your answer is 210. Thank you guys for watching. This was very easy question. If you have any doubts, please let me know in the comments below do not hesitate. Also, please like the video and subscribe the channel. Thank you so much. "

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Find the smallest number by which 200 should be multiplied to make it a perfect cube

We find 200 = `bar(2 xx 2 xx 2) xx 5 xx 5`

Grouping the prime factors of 200 as triplets, we are left with 5 × 5

We need one more 5 to make it a perfect cube.

So to make 200 a perfect cube multiply both sides by 5.

200 × 5 = `(bar(2 xx 2 xx 2) xx 5 xx 5) xx 5`

1000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5

Now 1000 is a perfect cube.

∴ The required number is 5.

Concept: Concept of Cube Root

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Which of the options below is the smallest number such that when multi [#permalink]

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From the options below, which is the smallest number such that when multiplied by 100! is divisible by 3^50?(A) 42(B) 72(C) 144(D) 216(E) 789 _________________

Intern

Joined: 10 Nov 2020

Posts: 26

Re: Which of the options below is the smallest number such that when multi [#permalink]

Considering 100!Taking first all powers of single 3.3,6,9,12...till 99. Thus one power of 3 will be 33.Next considering multiples of 9 since we need to take extra powers of 3 for that - 11 multiples. Thus total 33+11=44.Then taking power of 3 - 27,54,81 - another 3 taking total to 47.Then power of 4 - only 1 - 81. total 48. Thus we need 2 more for it to be divisible by 3^50.

Thus Answer is B. 72 (3^2*2^3)

Re: Which of the options below is the smallest number such that when multi [#permalink]

(X * 100 !) / 3^50 Highest power of 3 in 100 ! is 48. How ? 100/3 = 33, 33/3=11, 11/3=3, 3/3=1Total= 33+11+3+1= 48so we need at least two 3s to divide 100! from 3^50. Therefore, the correct option will be multiple of 9 (3*3)Check the options with the smallest number -

a) 42 is not a multiple of 9. Eliminate it.

b) 72 is a multiple of 9. Mark the option and move on. No need to check other options because we are looking the smallest number. Other options are higher than option B.

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Intern

Joined: 28 Dec 2020

Posts: 4

Re: Which of the options below is the smallest number such that when multi [#permalink]

So, lets find the power of 3 in 100! For that we will use the formula that we have [n/p^1] + [n/p^2] + [n/p^3] + [n/p^4] + ..... (continues till n>p)n= 100p = 3 >> [100/3^1] + [100/3^2] + [100/3^3]+ [100/3^4 ] = 33 +11+ 3+ 1 = 48 >>> 3^48 2) 3^48 / 3^50 = 1/3^2 = 1/9 (A) 42 * 1/9 => 42 is not divisible by 9 (B) 72 * 1/9 => 72 is divisble by 9

Answer is B

Re: Which of the options below is the smallest number such that when multi [#permalink]