Question

find the smallest five digit number which is exactly divisible by 32, 40, and 56. (solution with please)​

Answer 1

Answer:

1008 0

Step-by-step explanation:

Start by finding the least common multiple of those four numbers. One way to do this is to factorise each of them into prime numbers and pick the number which has all of the factors as many times as necessary

16 = 2×2×2×2

18 = 2×3×3

24 = 2×2×2×3

30 = 2×3×5

So we need four 2×2×2×2 for sixteen, another 3×3 for 18—these will be enough for 24—and a further 5 for 30. So the least common multiple is 2×2×2×2×3×3×5 = 720.

Now multiply 720 by the smallest number that results in 10,000 or greater. 720×14 = 10,080.

So the answer is 10,080, which is 16×630, 18×560, 24×420, or 30×336.

Answer 2

Given:

Smallest 5 digit number which is exactly divisible by 32, 40, and 56.

To find :

The smallest 5 digit number which is exactly divisible by 32, 40, and 56

Solution:

Step 1 of 2:

The smallest 5 digit number = 10000

L.C.M of 32, 40, 56 is,

32 = 2 × 2  × 2  × 2 × 2  

40 = 2 × 2  × 2  × 5  

56 = 2 × 2  × 2 × 7  

The L.C.M = 2 × 2 × 2 × 2 × 2 × 5 × 7

L.C.M = 1120

Divide the Smallest 5 digit number by 1120,  

When dividing 10000 by 1120 we get quotient as 8 and remainder as 1040.

Step 2 of 2:

The smallest 5 digit number which is exactly divisible by  32, 40, and 56​ is,

10000 + ( L.C.M - Remainder )

10000 + 1120 - 1040 = 10080

Final answer:

The smallest 5 digit number which is exactly divisible by 32, 40, and 56​ is 10080.