GSEB Class 6 Maths Solutions Chapter 3 સંખ્યા સાથે Exercise 3.3

Question 1. વિભાજ્યતાની ચાવીનો ઉપયોગ કરીને નીચેની કઈ સંખ્યા 2 વડે, 3 વડે, 4 વડે, 5 વડે, 6 વડે, 8 વડે, 9 વડે, 10 વડે અને 11 વડે વિભાજ્ય છે તે નક્કી કરોઃ
Answer:
Explanation: [Remember...]
For 2: A number is exactly divisible by 2 if its last digit is 0, 2, 4, 6, or 8.
For 3: A number is exactly divisible by 3 if the total of its digits can be exactly divided by 3.
For 4: A number is exactly divisible by 4 if the number formed by its last two digits (ones and tens place) can be exactly divided by 4.
For 5: A number is exactly divisible by 5 if its last digit is 0 or 5.
For 6: A number is exactly divisible by 6 if it is exactly divisible by both 2 and 3.
For 8: A number is exactly divisible by 8 if the number formed by its last three digits (hundreds, tens, and ones place) can be exactly divided by 8.
For 9: A number is exactly divisible by 9 if the total of its digits can be exactly divided by 9.
For 10: A number is exactly divisible by 10 if its last digit is 0.
For 11: A number is exactly divisible by 11 if the difference between the sum of the digits at odd places from the right and the sum of the digits at even places from the right is either 0 or a multiple of 11.

In simple words: To check divisibility, use the special rules for each number (2, 3, 4, etc.). For example, a number is divisible by 2 if its last digit is even. It's divisible by 3 if its digits add up to a multiple of 3. Apply these rules to each given number and mark 'હા' (Yes) if it's divisible, or 'ના' (No) if it's not.

Exam Tip: Memorize the divisibility rules thoroughly. For larger numbers, applying these simple rules can save a lot of time compared to actual division.

Question 2. વિભાજ્યતાની ચાવીનો ઉપયોગ કરીને નીચેનામાંથી કઈ સંખ્યા 4 અને 8 વડે વિભાજ્ય છે તે નક્કી કરોઃ
Answer:
Explanation:
(i) A number is perfectly divisible by 4 if the number formed by its last two digits can be perfectly divided by 4.
(ii) A number is perfectly divisible by 8 if the number formed by its last three digits can be perfectly divided by 8.

(a) 572.
(i) The number formed by the last two digits is 72. \( 72 \div 4 = 18 \).
\( \implies \) So, 572 is perfectly divisible by 4.
(ii) The number formed by the last three digits is 572. \( 572 \div 8 \) is not an exact division, as \( 572 \div 8 = 71.5 \).
\( \implies \) So, 572 is not perfectly divisible by 8.

(b) 726352
(i) The number formed by the last two digits is 52. \( 52 \div 4 = 13 \).
\( \implies \) So, 726352 is perfectly divisible by 4.
(ii) The number formed by the last three digits is 352. \( 352 \div 8 = 44 \).
\( \implies \) So, 726352 is perfectly divisible by 8.

(c) 5500
The number formed by the last two digits is 00. As 00 is divisible by 4, 5500 is perfectly divisible by 4.
The number formed by the last three digits is 500. As 500 is not perfectly divisible by 8, 5500 is not perfectly divisible by 8.

(d) 6000
6000 is perfectly divisible by 4.
6000 is perfectly divisible by 8.

(e) 12159
12159 is not perfectly divisible by 4.
12159 is not perfectly divisible by 8.

(f) 14560
14560 is perfectly divisible by 4.
14560 is perfectly divisible by 8.

(g) 21084
21084 is perfectly divisible by 4.
21084 is not perfectly divisible by 8.

(h) 31795072
31795072 is perfectly divisible by 4.
31795072 is perfectly divisible by 8.

(i) 1700
1700 is perfectly divisible by 4.
1700 is not perfectly divisible by 8.

(j) 2150
2150 is not perfectly divisible by 4.
2150 is not perfectly divisible by 8.
In simple words: To see if a number can be divided by 4, just check the last two digits. If those last two digits can be divided by 4, the whole number can be. For dividing by 8, look at the last three digits. If they can be divided by 8, then the entire number can be.

Exam Tip: When checking divisibility by 4 and 8, remember to isolate and test only the last two or three digits, respectively. This simplifies the process for very large numbers.

Question 3. વિભાજ્યતાની ચાવીનો ઉપયોગ કરીને નીચેનામાંથી કઈ સંખ્યા 6 વડે વિભાજ્ય છે તે નક્કી કરોઃ
Answer:
Explanation:
A number is completely divisible by 6 only if it can be completely divided by both 2 and 3.
To be fully divisible by 2, the last digit of the number should be 0, 2, 4, 6, or 8.
To be fully divisible by 3, the sum of the digits of the number should be fully divisible by 3.

(a) 297144
(i) The last digit of the number is 4.
\( \implies \) So, the number is perfectly divisible by 2.
(ii) The sum of the digits of the number is \( 2 + 9 + 7 + 1 + 4 + 4 = 27 \). 27 is perfectly divisible by 3.
\( \implies \) So, the number is perfectly divisible by 3.
Thus, the number is perfectly divisible by both 2 and 3.
\( \implies \) Therefore, 297144 is perfectly divisible by 6.

(b) 1258
(i) The last digit of the number is 8.
\( \implies \) So, the number is perfectly divisible by 2.
(ii) The sum of the digits of the number is \( 1 + 2 + 5 + 8 = 16 \). 16 is not perfectly divisible by 3.
\( \implies \) So, the number is not perfectly divisible by 3.
Thus, the number is divisible by 2 but not by 3.
\( \implies \) Therefore, 1258 is not perfectly divisible by 6.

(c) 4335
The last digit of the number is 5.
\( \implies \) So, the number is not perfectly divisible by 2.
Thus, there is no need to check for divisibility by 3.
\( \implies \) Therefore, 4335 is not perfectly divisible by 6.

(d) 61233
The last digit of the number is 3.
\( \implies \) So, the number is not perfectly divisible by 2.
\( \implies \) Therefore, 61233 is not perfectly divisible by 6.

(e) 901352
(i) The last digit of the number is 2.
\( \implies \) So, the number is perfectly divisible by 2.
(ii) The sum of the digits of the number is \( 2 + 5 + 3 + 1 + 0 + 9 = 20 \). 20 is not perfectly divisible by 3.
\( \implies \) So, the number is not perfectly divisible by 3.
\( \implies \) Therefore, 901352 is not perfectly divisible by 6.

(f) 438750
(i) The last digit of 438750 is 0.
\( \implies \) So, the number is perfectly divisible by 2.
(ii) The sum of the digits of the number is \( 0 + 5 + 7 + 8 + 3 + 4 = 27 \). 27 is perfectly divisible by 3.
\( \implies \) So, the number is perfectly divisible by 3.
\( \implies \) Therefore, 438750 is perfectly divisible by 6.

(g) 1790184
1790184 is perfectly divisible by 6. (Check as above.)

(h) 12583
12583 is not perfectly divisible by 6.
Note: 12583 is an odd number.

(i) 639210
639210 is perfectly divisible by 6. (Check as above.)

(j) 17852
17852 is not perfectly divisible by 6. (Check as above.)
In simple words: A number can only be divided by 6 if it can be divided by both 2 and 3. First, check if the last digit is even (0, 2, 4, 6, or 8) to see if it's divisible by 2. Then, add up all the digits in the number. If that sum can be divided by 3, then the number is also divisible by 3. If both of these things are true, then the original number is divisible by 6. If even one condition fails, it's not.

Exam Tip: Always remember that divisibility by 6 is a composite rule; both conditions (divisibility by 2 AND by 3) must be met for a number to be divisible by 6.

Question 4. વિભાજ્યતાની ચાવીનો ઉપયોગ કરીને નીચેનામાંથી કઈ સંખ્યા 11 વડે વિભાજ્ય છે તે નક્કી કરોઃ
Answer:
Explanation:
A number is exactly divisible by 11 if the difference between the sum of the digits at odd places from the right and the sum of the digits at even places from the right is either zero or a multiple of 11.

(a) 5445
The sum of the digits at odd places from the right of the number \( = 5 + 4 = 9 \).
The sum of the digits at even places from the right of the number \( = 4 + 5 = 9 \).
The difference between these two sums \( = 9 - 9 = 0 \).
This difference is 0.
\( \implies \) Therefore, 5445 is perfectly divisible by 11.

(b) 10824
The sum of the digits at odd places from the right of the number \( = 4 + 8 + 1 = 13 \).
The sum of the digits at even places from the right of the number \( = 2 + 0 = 2 \).
The difference between these two sums \( = 13 - 2 = 11 \).
This difference is a multiple of 11 ( \( 11 \times 1 = 11 \) ).
\( \implies \) Therefore, 10824 is perfectly divisible by 11.

(c) 7138965
The sum of the digits at odd places from the right of the number \( = 5 + 9 + 3 + 7 = 24 \).
The sum of the digits at even places from the right of the number \( = 6 + 8 + 1 = 15 \).
The difference between these two sums \( = 24 - 15 = 9 \).
Here, the difference is 9, which is neither 0 nor a multiple of 11.
\( \implies \) Therefore, 7138965 is not perfectly divisible by 11.

(d) 70169308
The sum of the digits at odd places from the right of the number \( = 8 + 3 + 6 + 0 = 17 \).
The sum of the digits at even places from the right of the number \( = 0 + 9 + 1 + 7 = 17 \).
The difference between these two sums \( = 17 - 17 = 0 \).
Here, the difference is 0.
\( \implies \) Therefore, 70169308 is perfectly divisible by 11.

(e) 10000001
The sum of the digits at odd places from the right of the number \( = 1 + 0 + 0 + 0 = 1 \).
The sum of the digits at even places from the right of the number \( = 0 + 0 + 0 + 1 = 1 \).
The difference between these two sums \( = 1 - 1 = 0 \).
Here, the difference is 0.
\( \implies \) Therefore, 10000001 is perfectly divisible by 11.

(f) 901153
The sum of the digits at odd places from the right of the number \( = 3 + 1 + 0 = 4 \).
The sum of the digits at even places from the right of the number \( = 5 + 1 + 9 = 15 \).
The difference between these two sums \( = 15 - 4 = 11 \).
Here, the difference is a multiple of 11 ( \( 11 \times 1 = 11 \) ).
\( \implies \) Therefore, 901153 is perfectly divisible by 11.
In simple words: To check if a number can be divided by 11, take the digits from the right. Add up the digits in the 1st, 3rd, 5th, etc., spots (odd places). Then, add up the digits in the 2nd, 4th, 6th, etc., spots (even places). Find the difference between these two sums. If this difference is 0 or can be divided by 11, then the original number is divisible by 11.

Exam Tip: Carefully identify odd and even places from the right (not from the left) when calculating the alternating sum for divisibility by 11. A common mistake is to start from the left.

Question 5. નીચે આપેલી સંખ્યાની દરેક ખાલી જગ્યામાં સૌથી નાનો અને સૌથી મોટો અંક લખો. જેથી તે સંખ્યાને 3 વડે ભાગી શકાય?
Answer:
Explanation: To solve this, we will find the sum of the digits in the incomplete number provided. We will then subtract this sum from the next highest multiple of 3 to discover the missing digit.

(a) .....6724
Let the missing digit be \( x \). The number is \( x6724 \).
The sum of the given digits is \( 6 + 7 + 2 + 4 = 19 \).
For the number to be divisible by 3, the sum of all its digits (\( x + 19 \)) must be a multiple of 3.
Multiples of 3 greater than or equal to 19 are 21, 24, 27, 30, and so on.

To find the smallest possible digit for \( x \):
If \( x + 19 = 21 \implies x = 21 - 19 \implies x = 2 \). This is a single digit.
So, the smallest digit is 2.

To find the largest possible digit for \( x \):
If \( x + 19 = 27 \implies x = 27 - 19 \implies x = 8 \). This is a single digit.
If \( x + 19 = 30 \implies x = 30 - 19 \implies x = 11 \). This is a two-digit number and is not valid.
So, the largest digit is 8.
Thus, the smallest digit is 2 and the largest digit is 8.

(b) 4765......2
Let the missing digit be \( x \). The number is \( 4765x2 \).
The sum of the given digits is \( 4 + 7 + 6 + 5 + 2 = 24 \).
For the number to be divisible by 3, the sum of all its digits (\( x + 24 \)) must be a multiple of 3.
Multiples of 3 greater than or equal to 24 are 24, 27, 30, 33, 36, and so on.

To find the smallest possible digit for \( x \):
If \( x + 24 = 24 \implies x = 24 - 24 \implies x = 0 \). This is a single digit.
So, the smallest digit is 0.

To find the largest possible digit for \( x \):
If \( x + 24 = 33 \implies x = 33 - 24 \implies x = 9 \). This is a single digit.
If \( x + 24 = 36 \implies x = 36 - 24 \implies x = 12 \). This is a two-digit number and is not valid.
So, the largest digit is 9.
Thus, the smallest digit is 0 and the largest digit is 9.
In simple words: To find the missing digit so a number can be divided by 3, first add up all the digits you have. Then, find the smallest and largest single digits that, when added to your sum, make a total that can be divided by 3. Remember, the missing digit must be between 0 and 9.

Exam Tip: When finding missing digits for divisibility by 3, always calculate the sum of existing digits and then determine which single digits (0-9) complete the sum to the nearest multiples of 3.

Question 6. નીચે આપેલી સંખ્યાની દરેક ખાલી જગ્યામાં સૌથી નાનો અને સૌથી મોટો અંક લખો. જેથી તે સંખ્યાને 11 વડે ભાગી શકાય?
Answer:
Explanation: A number is exactly divisible by 11 if the difference between the sum of its digits at odd places from the right and the sum of its digits at even places from the right is either zero or a multiple of 11.

(a) 92.....389
Let the missing digit be \( x \). The number is \( 92x389 \).
Sum of the digits at odd places from the right \( = 9 + 3 + 2 = 14 \).
Sum of the digits at even places from the right \( = 8 + x + 9 = 17 + x \).

Case (i): If the difference is 0.
\( (17 + x) - 14 = 0 \)
\( 3 + x = 0 \)
\( x = -3 \). This is not a valid digit (must be between 0 and 9).

Case (ii): If the difference is a multiple of 11.
Multiples of 11 are 11, 22, 33, and so on.

Consider the difference to be 11:
\( (17 + x) - 14 = 11 \)
\( 3 + x = 11 \)
\( x = 8 \). This is a valid single digit.

Consider the difference to be 22:
\( (17 + x) - 14 = 22 \)
\( 3 + x = 22 \)
\( x = 19 \). This is a two-digit number and is not valid.
Therefore, the only valid single digit is 8.
\( \implies \) The smallest and largest digit is 8.
Thus, the number is 928389.

(b) 8.....9484
Let the missing digit be \( x \). The number is \( 8x9484 \).
Sum of the digits at odd places from the right \( = 4 + 4 + x = 8 + x \).
Sum of the digits at even places from the right \( = 8 + 9 + 8 = 25 \).

Case (i): If the difference is 0.
\( 25 - (8 + x) = 0 \)
\( 17 - x = 0 \)
\( x = 17 \). This is a two-digit number and is not valid.

Case (ii): If the difference is a multiple of 11.
Multiples of 11 are 11, 22, 33, and so on.

Consider the difference to be 11:
\( 25 - (8 + x) = 11 \)
\( 17 - x = 11 \)
\( x = 6 \). This is a valid single digit.

Consider the difference to be 22:
\( 25 - (8 + x) = 22 \)
\( 17 - x = 22 \)
\( x = -5 \). This is not a valid digit.
Therefore, the only valid single digit is 6.
\( \implies \) The smallest and largest digit is 6.
Thus, the number is 869484.
In simple words: To find the missing digit for divisibility by 11, first calculate two sums: one for digits in odd positions from the right, and another for digits in even positions from the right. Then, find the difference between these sums. The missing digit must make this difference either 0 or a number that can be divided by 11. Pick the single digit (0-9) that works.

Exam Tip: When finding a missing digit for divisibility by 11, make sure to consider both possibilities for the difference (zero or a multiple of 11) and test only single-digit values (0-9) for the missing place.