Math, asked by sagarmohalkar2002, 9 months ago

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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Answered by knjroopa
5

Step-by-step explanation:

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.

Reference link will be

https://brainly.in/question/16332703

Answered by swethabhasker08
0

Answer:

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