### Playing With Numbers Class 8th Mathematics AP Board Solution

##### Question 1.Find the sum of integers which are divisible by 5 from 1 to 100.Answer:Given, Sum of integers from 1 to 100Need to find out the numbers which are divisible by 5.The numbers which are divisible by 5 are 5,10,15…..100⇒ Now, Sum of integers form 1 to N = Sum of integers from 1 to 100 divisible by 5= (5 + 10 + ….. + 100)= 5(1 + 2 + …. + 20)  = (5 × 10 × 21)= 1050Hence, the sum of integers which are divisible by 5 from 1 to 100 is 1050.Question 2.Find the sum of integers which are divisible by 2 from 11 to 50.Answer:Given, Sum of integers from 11 to 50we need to find out the numbers which are divisible by 2⇒ Now, Sum of integers form 1 to N = ⇒ Sum of integers divisible by 2 from 11 to 50 = Sum of integers divisible by 2 from 1 to 50 – Sum of integers divisible by 2 from 1 to 49= (2 + 4 + 6………. + 50) – (2 + 4 + 6…. + 10)= 2(1 + 2 + 3…… + 25)-2(1 + 2 + .. + 5) = (25 × 26)-(5 × 6)= 650 – 30= 620Hence, the sum of integers which are divisible by 2 from 11 to 50 is 620Question 3.Find the sum of integers which are divisible by 2 and 3 from 1 to 50.Answer:Given, Sum of integers from 1 to 50Need to find out the numbers which are divisible by 2 and 3.Here, since the number should be divisible by 2 and also 3 it is enough to find the sum of integers which are divisible by 6.∴ the sum of numbers which are divisible by 6 from 1 to 50 = 6,12,18…48⇒ Now, Sum of integers form 1 to N = = (6 + 12 + 18 + ….. + 48)= 6(1 + 2 + 3 + …. + 8) = 3(8 × 9)= 216Hence, the sum of integers from 1 to 50 which are divisible by 2 and 3 is 216.Question 4.(n3 − n) is divisible by 3. Explain the reason.Answer:Given, (n3 – n) is divisible by 3Here, (n3 – n) can be written as n(n2 -1)⇒ n(n2 -12)= n(n-1)(n + 1)= (n-1)n(n + 1)It is the product of three consecutive numbersThe product of three consecutive numbers is divisible by 3Example:Let n = 2Substitute in the above equation we get⇒ (2-1)(2)(2 + 1)= 1 × 2 × 3= 6 which is divisible by 3Hence, (n3 – n) is divisible by 3Question 5.Sum of ‘n’ odd number of consecutive numbers is divisible by ‘n’. Explain the reason.Answer:Given, sum of ‘n’ odd number of consecutive numbers is divisible by ‘n’.Sum of n consecutive odd numbers = = n(2n-1)∴ it is multiple of nHence, sum of n consecutive odd numbers is divisible by nQuestion 6.Is 111 + 211 + 311 + 411 divisible by 5? Explain.Answer:Given, 111 + 211 + 311 + 411We need to find out the given number is divisible by 5No, the given number cannot be divisible by 5Explanation:For a number to be divisible by 5, it should end with 0 or 5.Hence, the given number cannot be divisible by 5Question 7. Find the number of rectangles of the given figure?Answer:Given, a rectangle gridWe need to find out the number of rectangles in the figure⇒ 1 + 2 + 3 + 4 + 5 + 6 = 21Hence, 21 rectangles are there in the given figure.Question 8.Rahul’s father wants to deposit some amount of money every year on the day of Rahul’s birthday. On his 1st birth day Rs.100, on his 2nd birth day Rs.300, on his 3rd birth day Rs.600, on his 4th birthday Rs.1000 and so on. What is the amount deposited by his father on Rahul’s 15th birthday.Answer: Money deposited by Rahul's father is as follows:100, 300, 600, 1000They are in following series100 + 1 × 0, 100 + 2 × 100, 300 + 3 × 100, 600 + 4 × 100Multiplication numbers 1, 2, 3 & 4 represent the age Rahul is turning. The multiplication factor is previous year deposit.If we say, n is number of years, Rahul is turning and T1 = 100. Then the series can be re written as (T0 = 0)T1 + n × T0 , T2 = T1 + n × 100, T3 = T2 + n × 100 = T1 + (n-1) × 100 + n × 100T4 = T3 + n × 100 = T1 + (n - 2) × 100 + (n - 1) × 100 + n × 100....T15 = T14 + 15 × 100= 1 × 100 + 2 × 100 + 3 × 100 + ........+ 15 × 100All the terms after T1 are in AP where, nth term is sum of n terms of the AP wherea = 100 and d = 100 n = 15Hence T15 = = n(2a + (n-1)d)/2= 15(200 + 14 × 100)/2= 15 × 800= 12000Question 9.Find the sum of integers from 1 to 100 which are divisible by 2 or 5.Answer:Given, Sum of integers from 1 to 100Need to find out the numbers which are divisible by 2 or 5.⇒The numbers which are divisible by 2 are 2,4,6….100And the numbers which are divisible by 5 are 5,10,15…..100⇒ Sum numbers are repeated twice hence, we need to take the LCM of 2 and 5 which is 10∴ sum of numbers which are divisible by 10 are 10,20……100⇒ Now, Sum of integers form 1 to N = ∴ Sum of integers divisible by 2 or 5 from 1 to 100 = Sum of integers from 1 to 100 divisible by 2 + Sum of integers from 1 to 100 divisible by 5 – Sum of integers from 1 to 100 divisible by 10= (2 + 4….. + 100) + (5 + 10 + ….. + 100)-(10 + 20 + …… + 100)= 2(1 + 2 + … + 50) + 5(1 + 2 + …. + 20)-10(1 + 2 + ….. + 10)  = (50 × 51) + (5 × 10 × 21)-(5 × 10 × 11)= 2550 + 1050-550= 3050Hence, the sum of integers from 1 to 100 which are divisible by 2 or 5 is 3050Question 10.Find the sum of integers from 11 to 1000 which are divisible by 3.Answer:Given, Sum of integers from 11 to 1000we need to find out the numbers which are divisible by 3⇒ Now, Sum of integers form 1 to N = ⇒ Sum of integers divisible by 3 from 11 to 1000 = Sum of integers divisible by 3 from 1 to 1000 – Sum of integers divisible by 3 from 1 to 10= (3 + 6 + 9………. + 999) – (3 + 6 + …. + 9)= 3(1 + 2 + 3…… + 333)-3(1 + 2 + .. + 3)= = = 3(333(167)) – 3(6)= 3(55611)-18= 166833-18= 166815Hence, the sum of integers from 11 to 1000 which are divisible by 3 is 166815.

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