Each-of-the-digits-2-4-6-and-8-can-be-used-once-and-once-only-in-writing-a-four-digit-number-What-is-the-sum-of-all-such-numbers-that-are-divisible-by-11-

Tinku Tara June 3, 2023 Permutation and Combination 0 Comments

Question Number 137745 by bemath last updated on 06/Apr/21

Each of the digits 2, 4, 6, and 8 can be used once and once only in writing a four-digit number. What is the sum of all such numbers that are divisible by 11?

Answered by TheSupreme last updated on 06/Apr/21

ABCD is div for 11 if A+C−B−D=0 or A+C−B−D=11 A+C=B+D i: 8+2=4+6 is the only possible solution A=(2,8),(4,6) B=(4,6),(2,8) C=(2,8),(4,6) D=(4,6),(2,8) n=4×2=8 Σ_(i=1) ^8 1000A_i +100B_i +10C_i +D_i = =1000ΣA_i +100ΣB_i +10ΣC_i +ΣD_i ΣA_i =40 [2+2+4+4+6+6+8+8] =1111×40=44440

A B C D i s d i v f o r 11 i f

A + C - B - D = 0 o r A + C - B - D = 11

A + C = B + D

i : 8 + 2 = 4 + 6 i s t h e o n l y p o s s i b l e s o l u t i o n

A = (2, 8), (4, 6)

B = (4, 6), (2, 8)

C = (2, 8), (4, 6)

D = (4, 6), (2, 8)

n = 4 \times 2 = 8

\underset{i = 1}{\sum^{8}} 1000 A_{i} + 100 B_{i} + 10 C_{i} + D_{i} =

= 1000 Σ A_{i} + 100 Σ B_{i} + 10 Σ C_{i} + Σ D_{i}

Σ A_{i} = 40 [2 + 2 + 4 + 4 + 6 + 6 + 8 + 8]

= 1111 \times 40 = 44440

Commented by otchereabdullai@gmail.com last updated on 06/Apr/21

thank you sir!

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