Question

1.5. Find the sum of all 2-digit numbers that leave a remainder of 3 when divided by 4.​

Answer 1

Answer:

10 to 99

the no. exactly divisibe by 7

14,21,28……98.

the no. which leaves reminder of 3.

add 3 to each no. which are exactly divisible by 7

10,17,24,31…94

(7+3=10 two digit, 98+3=101 not a 2digit no.)

this is a A.P. a=10,d=7

an=a+(n-1)d

94=10+(n-1)7

n=13

sn = n/2(a+l)

=13/2(10+94)

676

All such numbers are of the form

Smallest value of

Largest value of (such that it is still a digit number)

What is the sum of all 2-digit numbers, which leave a remainder 3 when divided by 11?

What is the sum of all two digit numbers each of which leaves remainder 3 when divided by 5?

What is the smallest four digit number which when divided by 6 leaves reminder 5 and when divided by 5 leaves reminder 3?

FIRST NUMBER = 10 & LAST NUMBER 94

MEANS SERIES = 10, 17, 24, 31 , …… 94

THIS IS AN AP, FIRST TERM = 10 , LAST TERM = 94, DIFFERENCE = 7

LAST TERM = FIRST TERM +(N-1)*D

94 = 10 +(7–1)*7

84 = (N-1)*7

12 = (N-1)

N = 13

NOW SUM OF N TERMS OF AP = N/2(1ST TERM + LAST TERM) = 13/2* (10+94)

= 13 * 104/2

= 13*52

= 676 (ANSWER)

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The first number is 10 which when divided by 7 leaves a remainder of 3.

The last number is 94.

The sum is an AP whose first term is 17 and the last one is 94.

Tn = a+(n-1)d or

94 = 10 + (n-1)*7, or

84 = (n-1)*7, or

n-1 = 12, or

n = 13.

S12 = (13/2)[2*10+(13–1)*7]

= (13/2)[20+84]

= (13/2)[104]

= 676.

Answer : 676.