can you crack the code 79314 1 number is correct 95643 2 number are correct 57319 2 numbers are correct the right position sum of the numbers is equal the last. 2 numbers a+b+c+d+e= D 10+e
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Given can you crack the code 79314 1 number is correct 95643 2 number are correct 57319 2 numbers are correct the right position sum of the numbers is equal the last. 2 numbers a+b+c+d+e= D 10+e
- Now first number is 7 9 3 1 4
- Here we have 7 and in the wrong position
- Second number is 9 5 6 4 3
- According to question 2 numbers are right, so we have 5 is in wrong position and 6 in the right position.
- Third number will be 5 7 3 1 9.
- So here 2 numbers are in right position and so it will be 5 and 7
- Now given A + B + C + D + E = D x 10 + E
Now 57620 = 5 + 7 + 6 + 2 + 0 = 2 x 10 + 0
So 20 = 20
Also 57628 will be 5 + 7 + 6 + 2 + 8 = 2 x 10 + 8
28 = 28
Therefore either of the two numbers will be the secret code.
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https://brainly.in/question/16332703
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