Question

Find the sum of all the three digit numbers which are NOT divisible by 7.
Guys tmr is my exam pls help mehh!!!​

Answer 1

ANSWER:

• The sum = 424214

GIVEN:

• The number should be three digit and not divisible by 7

TO FIND:

• The sum of all three digit that are not divisible by 7

EXPLANATION:

Sum of all three digit that are not divisible by 7 = sum of all 3 digit numbers - sum of all three digit numbers divisible by 7

Three digit numbers = 100, 101, 102,........,999

sum of all 3 digit numbers:

${\boxed{\large{\bold{n = \dfrac{l - a}{d} + 1}}}}$

l = 999

d = 1

a = 100

$n = \dfrac{999 - 100}{1} + 1$

n = 899 + 1

n = 900

${\boxed{\large{\bold{S_n= \dfrac{n}{2} (2a + (n - 1)d)}}}}$

$S_{900} = \dfrac{900}{2} (2(100) + (900 - 1)1 )$

$S_{900} = 450(200 + 899)$

$S_{900} = 450(1099)$

$S_{900} = 494550$

Sum of all three numbers divisible by 7:

15 × 7 = 105 which is the smallest three digit number divisible by 7

996/7 gives 2 as a remiander. Hence 994 is the largest three digit number divisible by 7

Series = 105, 112, 119, ........994

${\boxed{\large{\bold{n = \dfrac{l - a}{d} + 1}}}}$

l = 994

a = 105

d = 7

$n = \dfrac{994 - 105}{7} + 1$

$n = \dfrac{889}{7} + 1$

n = 127 + 1

n = 128

${\boxed{\large{\bold{S_n= \dfrac{n}{2} (2a + (n - 1)d)}}}}$

$S_{128}= \dfrac{128}{2} (2(105) + (128 - 1)7)$

$S_{128}= 64(210 + (127)7)$

$S_{128}= 64(210 + 889)$

$S_{128}= 64(1099)$

$S_{128}= 70336$

$\text{Sum of all three digit that are not }$

$\text{divisible by 7 = S_{900} - S_{128} }$

$S_{900} - S_{128} = 494550 - 70336$

$S_{900} - S_{128} = 424214$

HENCE THE SUM OF ALL THREE DIGIT NUMBERS NOT DIVISIBLE BY 7 = 424214.

Answer 2

Answer:

didi

Step-by-step explanation:

i cant because my sister is beside only she will scold me...