find the least number which when divided by 12,16,24,and 36 leaves a remainder 7 in each case.
Find the prime factorization of each integer.
12 = 2² ∙ 3
16 = 2⁴
24 = 2³ ∙ 3
36 = 2² ∙ 3²
The least common multiple (LCM) is the smallest number which can be exactly divided by each of the given number.
The LCM will be the product of multiplying the highest power of each prime number together.
In this case LCM = 2⁴ ∙ 3² = 16 ∙ 9 = 144
LCM for 12,16,24,36 = 144
Add a reminder 7
LCM + a reminder 7 = 144 + 7 = 151
The least number which when divided by 12,16,24,and 36 leaves a remainder 7 = 151
Proof:
151 / 12 = ( 144 + 7 ) / 12 = 144 / 12 + 7 / 12 = 12 + 7 / 12
A remainder = 7
151 / 16 = ( 144 + 7 ) / 16 = 144 / 16 + 7 / 16 = 9 + 7 / 16
A remainder = 7
151 / 24 = ( 144 + 7 ) / 24= 144 / 24 + 7 / 24 = 6 + 7 / 24
A remainder = 7
151 / 36 = ( 144 + 7 ) / 36= 144 / 36 + 7 / 36 = 4 + 7 / 36
A remainder = 7
To find the least number that satisfies the given conditions, we need to find the least common multiple (LCM) of 12, 16, 24, and 36 and then add 7 to it.
First, let's find the LCM of the given numbers:
Step 1: Prime factorize each number:
12 = 2^2 * 3
16 = 2^4
24 = 2^3 * 3
36 = 2^2 * 3^2
Step 2: Determine the highest power for each prime factor:
2: highest power is 4
3: highest power is 2
Step 3: Multiply the prime factors with their highest powers:
LCM = 2^4 * 3^2 = 144
Now, we add 7 to the LCM to get the least number:
Least number = LCM + 7 = 144 + 7 = 151
Therefore, the least number that when divided by 12, 16, 24, and 36 leaves a remainder of 7 in each case is 151.