Question

Find the smallest number which when divided by 28 and 32 leaves remainder 8 and 12.

Answer 1

Hint: Use Euclid’s division lemma, Dividend = (Divisor \[\times \]Quotient) + Remainder. This helps you to find the number with respective remainders.

Complete step-by-step answer:
Let us assume the required number as N.
Given conditions are:
When \[\dfrac{N}{28}\] is done we get 8 as remainder.
When \[\dfrac{N}{32}\] is done we get 12 as remainder.
So, first we will find the smallest number let it be K, which when divided by 28 and 32 leaves remainder 0 in both the cases. The LCM of two numbers implies the same description.
K = LCM(28,32).
We find the LCM by using the prime factorisation method.
So by converting 28 and 32 into multiplication of prime numbers we get:
\[\begin{align}
  & 28={{2}^{2}}\times 7 \\
 & 32={{2}^{5}} \\
\end{align}\]
So, now we find the least common multiple (LCM), we get:
\[K={{2}^{5}}\times 7=224\]
Now, by Euclid’s division lemma,
Dividend = (Divisor \[\times \]Quotient) + Remainder (remainder < divisor)
Now, we will write 224 in terms of 28:
224 = 28 \[\times \]8+0
224 = 224+0
We need the remainder to be 8.
So, making remainder to be 8, we get:
224 = 216 + 8 ….. (1)
Now we need to write 216 in terms of 28,
Again by Euclid’s division lemma, we get:
216 = (28 \[\times \]7) + 20 ….. (2)
By substituting (2) in (1), we get:
224 = ((28 \[\times \]7) + 20) + 8
By simplifying, we get:
224-20 = (28 \[\times \]7) + 8
204 = 28\[\times \]7 + 8 ….. (3)
Now we need to write 224 in terms of 32.
Again by Euclid’s division lemma, we get:
224 = (32 \[\times \]7) + 0
We need the remainder to be 12.
By converting remainder to 12, we get:
224 = 212 + 12 ….. (4)
Now we need to write 212 in terms of 32,
Again by Euclid’s division lemma, we get:
212 = (32 \[\times \]6) + 20 ….. (5)
By substituting (5) in (4), we get:
224 = ((32\[\times \]6) + 20) + 12
By simplifying, we get:
224-20 = (32 \[\times \]6) + 12
204 = 32 \[\times \]6 + 12 ….. (6)
By equations (3) and (6):
204 = 28 \[\times \]7 + 8 = 32 \[\times \]6 + 12 …..(7)
Now we can see that the number 204 when divided by 28 and 32 leaves remainder 8 and 12.
\[\therefore\]The smallest number which when divided by 28 and 32 leaves remainder 8 and 12 is 204.
204 divided by 28 leaves 8 as remainder and when 204 divided by 32 leaves 12 as remainder.

Note: Students may confuse and think that it is two different questions and may find 2 numbers satisfying 28 and 32 conditions separately, but it is asking for one number satisfying both of them.