Question

 Find the smallest 6-digit number which is exactly divisible by 63.

Answer 1

Hint: Key rule: If a number is divisible by both 9 and 7, then the number is divisible by 63.

The divisibility rules of required numbers:

Divisibility rule of 9:

If the sum of digits of a number is divisible by 9, then the number is divisible by 9.

Divisibility rule of 7:

Double the last digit of the number and subtract it from the remaining number, if the result is divisible by 7, then the number is divisible by 7.

Complete step-by-step answer:

Step 1:

To find: the smallest 6-digit number which is exactly divisible by 63

We know, the smallest 6-digit number is 1,00,000 up to the place value lakhs.

Step 2:

Factorization of number

Prime Factorization of 63 $ = 1 \times 3 \times 3 \times 7$

Co-prime factors of 63 $ = (9,7)$

Key rule: If a number is divisible by both 9 and 7, then the number is divisible by 63.

Step 3:

The divisibility rules of required numbers

Divisibility rule of 9:

If the sum of digits of a number is divisible by 9, then the number is divisible by 9.

Divisibility rule of 7:

Double the last digit of the number and subtract it from the remaining number, if the result is divisible by 7, then the number is divisible by 7.

Step 4: 

CHECK DIVISIBILITIES

Check divisibility of 9 for 100000.

1+0+0+0+0+0+0 = 1

1 is not divisible by 9, thus 100000 is not divisible by 9.

Hence, 100000 is not divisible by 63.

If we add 8 to the 100000 

i.e. 100000 + 8 = 100008, the number will be divisible by 9.

Check for the divisibility of 7 for 100008:

\[{\text{    }}10000{\mathbf{8}}{\text{   }}\left( {{\text{8}} \times {\text{2}}} \right)\]

\[ \underline { - {\text{   }}16} \]

\[ {\text{ }}998{\mathbf{4}}{\text{   }}\left( {4 \times 2} \right)\]

\[ \underline { - {\text{   8}}} \]

\[ {\text{ 99}}{\mathbf{0}}{\text{   }}\left( {{\text{0}} \times {\text{2}}} \right)\]

\[ \underline { - {\text{ 0}}} \]

\[ {\text{ 99}}\]

99 is not divisible by 7. Thus, 100008 is not divisible by 7.

Hence, 100008 is not divisible by 63.

Next multiple of 9 = 100008 + 9

= 100017

Check for the divisibility for 7 for 100017:

\[{\text{    }}10001{\mathbf{7}}{\text{   }}\left( {7 \times {\text{2}}} \right)\]

\[ \underline { - {\text{   }}14} \]

\[ {\text{ }}999{\mathbf{7}}{\text{   }}\left( {7 \times 2} \right)\]

\[ \underline { - {\text{  14}}} \]

\[ {\text{ 98}}{\mathbf{5}}{\text{   }}\left( {5 \times {\text{2}}} \right)\]

\[ \underline { - 1{\text{0}}} \]

\[ {\text{ 88}}\]

88 is not divisible by 7. Thus, 100017 is not divisible by 7.

Hence, 100017 is not divisible by 63.

Next multiple of 9 = 100017 + 9

= 100026

Check for the divisibility of 7 for 100026:

\[{\text{    }}10002{\mathbf{6}}{\text{   }}\left( {6 \times {\text{2}}} \right)\]

\[ \underline { - {\text{   }}12} \]

\[  {\text{ }}999{\mathbf{0}}{\text{   }}\left( {0 \times 2} \right)\]

\[ \underline { - {\text{   0}}} \]

\[ {\text{ 99}}{\mathbf{9}}{\text{   }}\left( {9 \times {\text{2}}} \right)\]

\[ \underline { - 18} \]

\[ {\text{ 81}}\]

81 is not divisible by 7. Thus, 100026 is not divisible by 7.

Hence, 100026 is not divisible by 63.

Next multiple of 9 = 100026 + 9

= 100035

Check for the divisibility of 7 for 100035:

\[{\text{    }}10003{\mathbf{5}}{\text{   }}\left( {5 \times {\text{2}}} \right)\]

\[ \underline { - {\text{   }}10} \]

\[ {\text{ }}999{\mathbf{3}}{\text{   }}\left( {3 \times 2} \right)\]

\[ \underline { - {\text{   6}}} \]

\[ {\text{ 99}}{\mathbf{3}}{\text{   }}\left( {3 \times {\text{2}}} \right)\]

\[ \underline { - {\text{ 6}}} \]

\[ {\text{ 93}}\]

93 is not divisible by 7. Thus, 100035 is not divisible by 63.

Hence, 100035 is not divisible by 63.

Next multiple of 9 = 100035 + 9

= 100044

Check for the divisibility of 7 for 100044:

\[{\text{    }}10004{\mathbf{4}}{\text{   }}\left( {4 \times {\text{2}}} \right)\]

\[  \underline { - {\text{    8}}} \]

\[ {\text{ }}999{\mathbf{6}}{\text{   }}\left( {6 \times 2} \right)\]

\[ \underline { - {\text{ 12}}} \]

\[ {\text{ 98}}{\mathbf{7}}{\text{   }}\left( {7 \times {\text{2}}} \right)\]

\[ \underline { - 14} \]

\[ {\text{ 84}}\]

84 is divisible by 7. Thus, 100044 is divisible by 7.

As 100044 is divisible by both the co-prime factors of 63, the number is divisible by 63.

100044 is the smallest 6-digit number which is exactly divisible by 63.

Note: Students may go wrong while checking divisibility by 7 as they forget to remove unit’s digit while subtracting. For example- while dividing 100008 by 7, first double unit’s number 8 and then subtract 16 from 10000 not 100008. This may confuse and create problems for students.

Well, there is a short trick, the step we have done to make 100000 divisible by 9. 

On dividing $\dfrac{{100000}}{9}$, the remainder is 1

Either subtract 1 from 100000 or add 8 (9-1) for the number exactly divisible by 9.

But the condition is the number should be 6-digit.

$\therefore $8 is added to 100000. Thus 100008 is divisible by 9.

Similarly, on dividing $\dfrac{{100000}}{{63}}$remainder is 19.

Either subtract 19 from 100000 or add 44 (63 - 19) for the number exactly divisible by 63.

But the condition is the number should be 6-digit.

$\therefore $44 is added to 100000. Thus 100044 is divisible by 63.