I am a 4-digit number divisible by 3. My tens digit is three times as great as my thousands digit, and the sum of my digits is 15. If you reverse my digit, I am divisible by 2 as well as by 6. What am I?
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Answers
Step-by-step explanation:
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Solution :-
Let us assume that, the four digit number is ABCD where ,
- A = Thousand's digit
- B = Hundred's digit
- C = Ten's digit
- D = Unit digit .
Statement 1) :- If you reverse my digit, I am divisible by 2 as well as by 6.
Conclusion :-
- Reversed number = DCBA
- Divisible by 2 , so DCBA must be a even number . Then , A = 2, 4, 6 or 8 . { if we take 0 also ABCD will not be a four digit number . }
- Already divisible by 3 since sum is equal to 15 .
- Then , the number will also be divisible by 6 .
Statement 2) :- My tens digit is three times as great as my thousands digit .
Conclusion :-
- C = 3A
- from first statement conclusion A = 2,4, 6 or 8 .
- Only possible value of A we can take now is 2 as because if we take 4, 6 and 8 , C becomes a two digit number which is not possible .
- So, A = 2, then C = 6 .
Statement 3) :- Sum of my digits is 15 .
Conclusion :-
→ A + B + C + D = 15
→ 2 + B + 6 + D = 15
→ 8 + B + D = 15
→ B + D = 15 - 8
→ B + D = 7
So, Possible values of B and D are :-
- When B = 0, D = 7
- When B = 1, D = 6
- When B = 2, D = 5
- When B = 3, D = 4
- When B = 4, D = 3
- When B = 5, D = 2
- When B = 6, D = 1
- When B = 7, D = 0
Putting values of A, B, C and D we get, required 4 - digit number can be :-
- 2067
- 2166
- 2265
- 2364
- 2463
- 2562
- 2661
- 2760
Therefore, required 4 - digits numbers will be :- 2067
2067 2166, 2265, 2364, 2463, 2562, 2661 and 2760 .
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