please give me this question answer I will do Mark as branilst
Answers
Answer:
PLEASE MARK AS BRAINLIEST ANSWER.
Step-by-step explanation:
→ (a) = Answer in attachment.
→ (b) = Answer in attachment.
→ (d) = Answer in attachment.
→ (c) = Given : 4-digits numbers using each the digits 1,2,4 and 5 such that the numbers made are divisible by 132
→ To find : Two such numbers
→ Solution
132 = 2 * 2 * 3 * 11
=> 132 = 4 * 3 * 11
→ Divisible by 2 so last digit should be even 2 or 4
a number is divisible by 4 if sum of two digits are divisible by 4
last two possible digits are
→ 12 , 52 , 24
→ 1 + 2 + 4 + 5 = 12 sum is divisible by 3
→ So now remaining divisibility by 11
→ Subtract the last digit from a number made by the other digits. If that number is divisible by 11 then the original number is, too.
→ Ending with 12
→ 5412 & 4512
→ 541 - 12 = 539 then 53 - 9 = 44 divisible by 11
→ 5412 is divisible by 132
→ 5412/132 = 41
→ 4512
→ 451 - 2 = 449 then 44 - 9 = 35 not divisible by 11
→ Ending with 52
→ 1452 & 4152
→ 145 - 2 = 143 then 14 - 3 = 11 divisible by 11
→ 1452 is divisible by 132
→ 1452/132 = 11
→ 4152
→ 415 - 2 = 413 then 41 - 3 = 38 not divisible by 11
→ Ending with
→ 1524 & 5124
→ 152 - 4 = 148 then 14 - 8 = 6 , not divisible by 11
→ 5124
→ 512 - 4 = 508 then 50 - 8 = 42 not divisible by 11
→ 1452 & 5412 are two 4-digits numbers using each the digits 1,2,4 and 5 such that the numbers made are divisible by 132
