Remove the digit

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Let $n$ be a five digit number ,whose first digit is non-zero, and let $m$ be the four digit number formed from n by removing its middle digit. Determine all $n$ such that $n/m$ is an integer

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Let ee0d5015270479d9e8179e1ea730b0b5f66109e7 and 08a16f67c0d03a834c87e9a0fde5f4e21d60ec44, where c7d457e388298246adb06c587bccd419ea67f7e88136a7ef6a03334a7246df9097e5bcc31ba33fd23372c1cb6d68cf97c2d231acc0b47b95a9ed04cc96ab646de7704969b91c76a214126b45f2b07b25, and a3a59bb1293ee3f6dec19de4019a7178874219ae are base-10 digits and 1cae493c4855ca1cdd17c420fc6a97ebc082b591. If ac3f8f77f019433405ef87fc8759c7df8e1177ef is an integer, then 6b3d61b25b562ac73cc48df6d09cf6100f6059d4, or

e5173ba6d2d7aab087071ff6af45804039ddd30c

This implies that

5db17f1c9e8fc33765fb8f7c4625dbdb8e736cfa

Clearly we have that 4ed434a0fd66602e4e80071550e2ca191d6beaad, as c7d457e388298246adb06c587bccd419ea67f7e8 is positive. Therefore, this quotient must be equal to 9 (note that this does not mean a6d4e58f56d2598145abaf316c85b829c5a77f69), and

c6f8738dfe2afd857f435686659b68a89aa5697e

This simplifies to e0dad51dc29af8dc182e6fa09692aa5ac1694ab6. The only way that this could happen is that a80d7a3ca845e6cae14b9db710b64331219ec3f7. Then 95aa2e1f2b723a532cbfd9a3e627f00907c43c98. Therefore the only values of 174fadd07fd54c9afe288e96558c92e0c1da733a such that ac3f8f77f019433405ef87fc8759c7df8e1177ef is an integer are multiples of 1000. It is not hard to show that these are all acceptable values.

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