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a. Find the least number such that when 5 is added, the sum is exactly divisible by 32, 64, and 192.

Sagot :

To solve this problem, we need to find the least number such that when 5 is added to it, the resulting sum is divisible by 32, 64, and 192. Here's a step-by-step solution:

1. Determine the Least Common Multiple (LCM):
To ensure that the resulting sum is divisible by 32, 64, and 192, we first need to find the Least Common Multiple (LCM) of these three numbers:

- The LCM of 32 and 64: Since 64 is a multiple of 32, the LCM is 64.
- The LCM of 64 and 192: Since 192 is a multiple of 64, the LCM is 192.

Therefore, the LCM of 32, 64, and 192 is 192.

2. Form the Equation:
Let [tex]\( x \)[/tex] be the least number we are looking for. According to the problem, when 5 is added to [tex]\( x \)[/tex], the sum must be divisible by 192. This can be mathematically represented as:
[tex]\[ x + 5 = 192k \][/tex]
for some integer [tex]\( k \)[/tex].

3. Isolate [tex]\( x \)[/tex]:
To find [tex]\( x \)[/tex], rearrange the equation:
[tex]\[ x = 192k - 5 \][/tex]

4. Find the Least Value of [tex]\( x \)[/tex]:
Since we want the least number [tex]\( x \)[/tex], we need to find the smallest value of [tex]\( k \)[/tex] such that [tex]\( x \)[/tex] is non-negative and as small as possible. The smallest integer [tex]\( k \)[/tex] is 1:
[tex]\[ x = 192 \times 1 - 5 \][/tex]
[tex]\[ x = 192 - 5 \][/tex]
[tex]\[ x = 187 \][/tex]

Hence, the least number with which when 5 is added to it the resulting sum is exactly divisible by 32, 64, and 192 is:

[tex]\[ \boxed{187} \][/tex]