Explore number patterns, puzzles, and interesting arithmetic tricks. These Class 8 Maths Ganita Prakash NCERT Solutions guide students to solve problems efficiently while enhancing number sense.
Figure it Out – Page 122
1. The sum of four consecutive numbers is 34. What are these numbers?
Solution:
Let the four consecutive numbers be x,x+1,x+2,x+3.
x+(x+1)+(x+2)+(x+3)=34 ⟹ 4x+6=34 ⟹ x=7
Hence, the numbers are 7, 8, 9, 10.
2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Solution:
If p is the largest of five consecutive numbers, the remaining four are (p−1),(p−2),(p−3), and (p−4).
3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
(i) The sum of two even numbers is a multiple of 3.
Solution:
Let the two even numbers be 2a and 2b.
Sum=2a+2b=2(a+b)
For 2(a+b) to be divisible by 3, (a+b) must be a multiple of 3.
Examples:
- 2+4=6 ⇒ divisible by 3
- 2+8=10 ⇒ not divisible by 3
Conclusion: The statement is sometimes true.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Solution:
If a number is divisible by 18, then it must also be divisible by 9, since 9 is a factor of 18.
18a÷9=2a⇒divisible by 9.
However, if a number is divisible by 9, it is not necessarily divisible by 18.
9b÷18=b2⇒may not be an integer.
Examples:
- 9 is divisible by 9 but not by 18.
- 27 is divisible by 9 but not by 18.
Conclusion: The statement is sometimes true.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Solution:
Let the two numbers be a and b.
If neither is divisible by 6, their sum may still be divisible by 6.
Example: 2 and 4 are not divisible by 6, but 2+4=6 is divisible by 6.
Conclusion: Sometimes true.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Solution:
Let a multiple of 6 be 6a and a multiple of 9 be 9b.
6a+9b=3(2a+3b) ⇒ always divisible by 3.
Examples:
- 6+9=15 → divisible by 3
- 12+18=30 → divisible by 3
Conclusion: Always true.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Solution:
Let a multiple of 6 be 6a and a multiple of 3 be 3b.
6a+3b=3(2a+b)
For this sum to be divisible by 9, (2a+b) must be a multiple of 3.
Examples:
- 6+3=9 → divisible by 9
- 6+6=12 → not divisible by 9
Conclusion: Sometimes true.
4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Solution:
The L.C.M. of 3 and 4 is 12.
Such numbers can be expressed as:
12a+2
Examples:
(i) 12×1+2=14
(ii) 12×2+2=26
(iii) 12×3+2=38
5. “I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around,
But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”
Solution:
When grouped in 3’s, 2’s, or 5’s, one pebble is left each time.
When grouped in 7’s, none is left.
Also, the number is ≤ 100.
LCM of 2,3,5=30
So, the number of pebbles must be of the form 30k+1.
- For k=1: 30+1=31 → not divisible by 7
- For k=2: 60+1=61 → not divisible by 7
- For k=3: 90+1=91 → divisible by 7
Hence, the number of pebbles is 91.
6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?
Solution:
A number that leaves a remainder of 2 when divided by 6 can be expressed as 6k+2.
Let three such numbers be (6a+2),(6b+2),(6c+2).
(6a+2)+(6b+2)+(6c+2)=6(a+b+c)+6=6(a+b+c+1)
This is clearly divisible by 6.
So, Tathagat’s claim is always true.
Examples:
- 20+26+32=78 → divisible by 6
- 2+8+14=24 → divisible by 6
7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661 (ii) 4779 – 661
Solution:
(i) 4779+661:
- Remainders → 5+3=8
- 8÷7 leaves remainder 1.
(ii) 4779−661:
- Remainders → 5−3=2
- So, remainder = 2.
8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Solution:
A number leaving remainder 2 when divided by 3 can be written as 3x+2.
A number leaving remainder 3 when divided by 4 is 4x+3.
A number leaving remainder 4 when divided by 5 is 5x+4.
Since the L.C.M. of 3, 4, and 5 is 60, and in each case the remainder is 1 less than the divisor, the required number must be 1 less than 60.
60−1=59
Thus, the smallest number satisfying all the conditions is 59.