Problem: Find the smallest number of 5 digits which when divided by 20, 24, 35, and 45 leaves 5 as remainder.

Solution:

To find the smallest number of 5 digits that satisfies the given conditions, we can use the method of finding the least common multiple (LCM) of the given numbers. The LCM will be the smallest number that is divisible by all the given numbers.

Finding the LCM:

We have four numbers: 20, 24, 35, and 45. Let's find their LCM step by step.

Step 1: Prime Factorization

Prime factorize each number to its prime factors:

20 = 2 * 2 * 5
24 = 2 * 2 * 2 * 3
35 = 5 * 7
45 = 3 * 3 * 5

Step 2: Identify Common and Uncommon Prime Factors

Now, identify the common and uncommon prime factors among the given numbers:

Common Prime Factors: 2, 3, 5
Uncommon Prime Factors: 2, 2, 5, 7, 3, 3

Step 3: Multiplication of Prime Factors

Multiply the common prime factors and uncommon prime factors separately:

Common Prime Factors: 2 * 3 * 5 = 30
Uncommon Prime Factors: 2 * 2 * 5 * 7 * 3 * 3 = 2520

Step 4: LCM Calculation

Now, calculate the LCM by multiplying the common prime factors and the uncommon prime factors:

LCM = Common Prime Factors * Uncommon Prime Factors
= 30 * 2520
= 75600

Smallest Number with Remainder 5:

To find the smallest number of 5 digits that leaves a remainder of 5 when divided by 20, 24, 35, and 45, we need to find the smallest multiple of the LCM (75600) that satisfies this condition.

Step 1: Starting from the LCM (75600), add 5 to it.

75600 + 5 = 75605

Step 2: Check if the number is divisible by 20, 24, 35, and 45.

75605 is not divisible by 20, 24, 35, and 45.

Step 3: Continue adding 5 until we find a number that satisfies the condition.

75605 + 5 = 75610
75610 + 5 = 75615
...
75655 + 5 = 75660

Step 4: Check if the number is divisible by 20, 24, 35, and 45.

75660 is divisible by 20, 24, 35, and 45.

Therefore, the smallest number of 5 digits that leaves a remainder of 5 when divided by 20, 24, 35, and 45 is 75660.

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