What is the smallest number which when divided by 54 36 and 27 leaves remainder of 31 13 and 4

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  • First find the LCM of 16,24,36 and 54. LCM=432. Smallest 5 - digit no.=10000 Divide 10000 by 432. Remainder=64 Req. no. = 10000+(432-64) =10000+368 =10368 Ans.=10368 is the smallest 5 digit no. exactly divisible by 16,24,36 and 54.
  • Find the smallest 4 digit number which when divided by 28, 36, 54 gives a remainder of 7 in each case.
  • Answer (Detailed Solution Below)
  • What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?
  • What is the smallest number which when divided by 24,36 and 54 gives a remainder of 5 each time?
  • Find the smallest number which when divided by 48 and 60, leaves a remainder 7 in each case but when divided by 13 leaves no remainder. Find the sum of the digits in this required smallest number.
  • Answer (Detailed Solution Below)
  • Detailed Solution
  • More LCM and HCF Questions
  • More Number System Questions
  • What is the smallest natural number which when divided by 24, 18, 40 and 60 leave remainder 8 in each case?
  • What is the smallest number to divide 24 and 36?
  • What is the smallest number which when divided by 24 30 and 54 leaves a remainder of 5 in each case?
  • What is the smallest number which when divided by 27 36 57 gives a remainder of 5 in each case?
  • What is the smallest number that when divided by 36 54 and 72 leaves remainders of 6 in each case?
  • What least number when divided by 36 24 and 16 leaves 11 as remainder in each case ?( Hint find LCM then add 11 to LCM?

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Solution

First find the LCM of 16,24,36 and 54. LCM=432. Smallest 5 - digit no.=10000 Divide 10000 by 432. Remainder=64 Req. no. = 10000+(432-64) =10000+368 =10368 Ans.=10368 is the smallest 5 digit no. exactly divisible by 16,24,36 and 54.

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Q.

Find the least 5-digit number which is exactly divisible by 20, 25, 30.

Find the smallest 4 digit number which when divided by 28, 36, 54 gives a remainder of 7 in each case.

  1. 1619
  2. 1520
  3. 1620
  4. 1519

Answer (Detailed Solution Below)

Option 4 : 1519

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Given:

The smallest 4 digit number then divide by 28, 36, 54 gives a remainder as 7.

Calculation:

LCM of (28,36,54) + 7 will be the number that we have to find

LCM of (28,36,54) will be

28 = 4 × 7

36 = 4 × 9 

54 = 6 × 9

LCM will be = 9 × 4 × 3 × 7 = 756

But we have to find the smallest 4 digit number

⇒ 756 × 2 = 1512 + 7

The smallest number is 1519.

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Guys, does anyone know the answer?

get find the smallest number which when divided by 24, 30, 48, and 60 leaves a remainder 2 in each case from screen.

What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?

Click here👆to get an answer to your question ✍️ What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?

What is the smallest number which when divided by 54 36 and 27 leaves remainder of 31 13 and 4

Question

What is the smallest number which when divided by 24,36 and 54 gives a remainder of 5 each time?

Hard Open in App Solution Verified by Toppr Let's factorize, 24 = 2x2x2x3 36 = 2x2x3x3 54 = 2x3x3x3

The smallest number divisible by all 24, 36, 54 is:

LCM(24, 36, 54) = 216

Thus, to get 5 as remainder, we need to add 5 to 216.

(as 216 is the smallest number which gives remainder 0 when divided by 24 or 36 or 54)

216 + 5 = 221

For 12, 221 gives 18 as quotient and 5 as remainder

For 36, 221 gives 6 as quotient and 5 as remainder

For 54, 221 gives 4 as quotient and 5 as remainder

Hence, 221 is required number.

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Given∶ A number when divided by 48 and 60, leaves a remainder 7 in each case. But when divided by 13 leaves no remainder. Formula Used∶ Concept of LCM a

Home Quantitative Aptitude Number System LCM and HCF LCM

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Find the smallest number which when divided by 48 and 60, leaves a remainder 7 in each case but when divided by 13 leaves no remainder. Find the sum of the digits in this required smallest number.

13 14 15 11

Answer (Detailed Solution Below)

Option 1 : 13

Detailed Solution

Download Solution PDF

Given

A number when divided by 48 and 60, leaves a remainder 7 in each case.

But when divided by 13 leaves no remainder.

Formula Used

Concept of LCM and Hit and Trial method

Calculation

LCM of 48 and 60 = 24 × 3 × 5 = 240

Let the required number be N

So, N = 240 × x + 7

Now use Hit and Trial method

When, x = 1

Then, N = 240 × 1 + 7 = 240 + 7 = 247

Hence, sum of the digits = 2 + 4 + 7 = 13

∴ 13 is the sum of the digits in the required smallest number which when divided by 48 and 60 leaves a remainder 7, but when divided by 13, the remainder is zero.

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स्रोत : testbook.com

What is the smallest natural number which when divided by 24, 18, 40 and 60 leave remainder 8 in each case?

Answer (1 of 6): LCM is the least no. /Smallest no. Which is divisible by all the given numbers. So, LCM of 24= 2× 2×2×3, 18= 2×3×3, 40= 2×2×2×5, 60=2×2×3×5, is all factors with greatest power, so LCM = 2^3 ×3^2 ×5^1= 8×9× 5 = 360. So , no. Leaves remainder 8 in each case is 360+8 = 368.

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What is the smallest natural number which when divided by 24, 18, 40 and 60 leave remainder 8 in each case?

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6 Answers HsBadarinath

Ph.D. in Civil Engineering. Maths keeps one mentally active.Upvoted by

Krogi Us

, M.Sc. Mathematics & Computer Science, Aalto University (1988)Author has 52.3K answers and 32.9M answer viewsUpdated 4y

Factors of 24 = 2x2x2x3 18 = 2x3x3 40 = 2x2x2x5 60 = 2x2x3x5

LCM = 2x2x2x3x3x5 = 360

360+8 = 368. Answer is 368. Related questions

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What is the lowest natural number which when divided by 14, 24, 2 and 10 leaves the remainder of 1 in each case?

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Ali Dursen

M.S. in Mathematics, Sabanci University (Graduated 2013)Upvoted by

Michael Jørgensen

, PhD in mathematicsAuthor has 220 answers and 135.7K answer views4y

Such a number would be 8 greater than a number that is divisible by all those numbers. Since we want the smallest, answer is 0+8.

TARUN SHARMA

Works at MathematicsUpvoted by

Krogi Us

, M.Sc. Mathematics & Computer Science, Aalto University (1988)4y

LCM is the least no. /Smallest no. Which is divisible by all the given numbers. So, LCM of 24= 2× 2×2×3, 18= 2×3×3, 40= 2×2×2×5, 60=2×2×3×5, is all factors with greatest power, so LCM = 2^3 ×3^2 ×5^1= 8×9× 5 = 360. So , no. Leaves remainder 8 in each case is 360+8 = 368.

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prime factors of 24,18,40 and 60 are:

24=2^3*3 18=2*3^2 40=2^3*5 60=2^2*3*5

Thus L;CM (24,18,40,60)=2^3 *3^2*5

=8*9*5=360

THE REMAINDER IN ALL CASES IS 8

HENCE THE NUMBER=LCM+8=368

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Using the J programming language:

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24= 2 3 ×318=2× 3 2 40= 2 3 ×560= 2 2 ×3×5

24=23×318=2×3240=23×560=22×3×5

L.C.M. (24, 18, 40, 60) =

2 3 × 3 2 ×5=360

L.C.M. (24, 18, 40, 60) =23×32×5=360

⟹360 + 8 = 368 is the smallest number which when divided by

⟹360 + 8 = 368 is the smallest number which when divided by

24, 18, 40 and 60 leave remainder 8 in each case.

24, 18, 40 and 60 leave remainder 8 in each case.

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स्रोत : www.quora.com

What is the smallest number to divide 24 and 36?

The LCM of two non-zero integers, 24 and 36, is the smallest positive integer 72 which is divisible by both 24 and 36 with no remainder.

What is the smallest number which when divided by 24 30 and 54 leaves a remainder of 5 in each case?

Step-by-step explanation: LCM(24, 30, 54)= 2*2*2*3*3*3*5=1080.

What is the smallest number which when divided by 27 36 57 gives a remainder of 5 in each case?

Answer. Answer: The smaller number which 27,36 and 57 is 18. The reminder of 5 in each case is 113.

What is the smallest number that when divided by 36 54 and 72 leaves remainders of 6 in each case?

Answer: LCM of 36, 54, and 72 is 216. Explanation: The LCM of three non-zero integers, a(36), b(54), and c(72), is the smallest positive integer m(216) that is divisible by a(36), b(54), and c(72) without any remainder.

What least number when divided by 36 24 and 16 leaves 11 as remainder in each case ?( Hint find LCM then add 11 to LCM?

This is Expert Verified Answer 155 is the least number when divided by 36, 24 and 16 leaves 11 as remainder in each case. 144 is the LCM of 36, 24 and 16. ∴ 155 is the least number when divided by 36, 24 and 16 leaves 11 as remainder in each case.

What is the smallest number which when divided by 24 36 and 54 leaves remainder of 5 each time?

Therefore, the smallest number which when divided by 24, 26 and 54 gives a remainder of 5 each time is 221.

What is the smallest number which when divided by 27 36 and 57 gives a remainder of 5 in each case?

Answer: The smaller number which 27,36 and 57 is 18. The reminder of 5 in each case is 113.

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The LCM of 27 and 36 is 108. To find the LCM of 27 and 36, we need to find the multiples of 27 and 36 (multiples of 27 = 27, 54, 81, 108; multiples of 36 = 36, 72, 108, 144) and choose the smallest multiple that is exactly divisible by 27 and 36, i.e., 108.

What least number when divided by 36 24 and 16 leaves 11 as remainder in each case ?( Hint find LCM then add 11 to LCM?

This is Expert Verified Answer 155 is the least number when divided by 36, 24 and 16 leaves 11 as remainder in each case. 144 is the LCM of 36, 24 and 16. ∴ 155 is the least number when divided by 36, 24 and 16 leaves 11 as remainder in each case.