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Asked by Praveen Last Modified
Answer To find the smallest number that leaves remainders of 8 when divided by 28 and 12 when divided by 32, we can use the Chinese Remainder Theorem (CRT).
The CRT states that if we have a system of congruences x≡ai(modmi)x≡ai(modmi) for i=1,2,…,ni=1,2,…,n, where the mimi are pairwise coprime, then there exists a unique solution xx modulo M=m1⋅m2⋅…⋅mnM=m1⋅m2⋅…⋅mn.
In our case, we have:
First, let's find the value of M=28×32=896M=28×32=896.
Next, we find the multiplicative inverses of 32 modulo 28 and of 28 modulo 32. Let's call these inverses y1y1 and y2y2 respectively.
y1y1 is such that 32×y1≡1(mod28)32×y1≡1(mod28). y2y2 is such that 28×y2≡1(mod32)28×y2≡1(mod32).
Using the Extended Euclidean Algorithm or observation, we find y1=22y1=22 and y2=9y2=9.
Now, we can use these inverses to find the solution:
x=(8×32×9+12×28×22)(mod896)x=(8×32×9+12×28×22)(mod896)
Let's compute this:
x=(2304+7392)(mod896)x=(2304+7392)(mod896) x=9696(mod896)x=9696(mod896) x=48x=48
So, the smallest number that satisfies the conditions is 48.
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