Question

Divided by 9, i leave a remainder of 6

Answer 1

Answer:

54

Step-by-step explanation:

because x/9=6

x=6x9

x=54

Answer 2

The number is 33.

When divided by 9, leaves remainder of 6, when divided by 4, leaves remainder of 1 and when divided by 10, leaves a reminder of 3.

Complete Question:

Given:

• When divided by 9, I leave the remainder of 6, when divided by 4, I leave the remainder of 1 and when divided by 10, I leave a reminder of 3. Who am I?

To find:

• Find the number.

Solution:

We know that according to Euclid's division algorithm, if 'a' and 'b' are two numbers, such that a>b, then

a=bq+r; 0≤r<b

here

q is quotient and r is the remainder.

Step 1:

Let the number be x.

When x is divided by 9, it leaves a reminder 6.

As we don't know quotient, so let it is q.

According to Euclid's Division lemma

$\bf x=9q+6...eq1$

Step 2:

When x is divided by 4, it leaves a reminder 1.

As we don't know the quotient, so let it be p.

According to Euclid's Division lemma

$\bf x=4p+1...eq2$

Step 3:

When x is divided by 10, it leaves a reminder 3.

As we don't know the quotient, so let it be m.

According to Euclid's Division lemma

$\bf x=10m+3...eq3$

Step 4:

Simplify eq1, eq2 and eq3, to find the value of x.

From eq1 and eq2.

$9q+6=4p+1$

$\implies \bf 4p=9q+5...eq4$

From eq2 and eq3.

$4p+1=10m+3$

$\implies \bf 4p=10m+2...eq5$

From eq4 and eq5

$9q+5=10m+2$

$\implies \bf 9q=10m-3...eq6$

Now, if we put m=1 and 2 in eq6, we will not get a number which is completely divisible by 9.

So, put m= 3

$9q=30-3$

$\bf q=3$

and when we put m= 3 in eq5

$4p=30+2$

$\bf p= 8$

So, put the values of p, q and m in eqs1, 2 and 3.

We will get x= 33.

Thus,

The number is 33,

which when divided by 9, leaves remainder of 6, when divided by 4, leaves remainder of 1 and when divided by 10, leaves a reminder of 3.

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