Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 110587 by Aina Samuel Temidayo last updated on 29/Aug/20

A palindrome is a number that  remains the same when its numbers  are reversed. The number n and  n+192 are three−digit and  four−digit palindromes respectively.  What is the sum of the digits of m?     Can this be solved mathematically?

$$\mathrm{A}\:\mathrm{palindrome}\:\mathrm{is}\:\mathrm{a}\:\mathrm{number}\:\mathrm{that} \\ $$$$\mathrm{remains}\:\mathrm{the}\:\mathrm{same}\:\mathrm{when}\:\mathrm{its}\:\mathrm{numbers} \\ $$$$\mathrm{are}\:\mathrm{reversed}.\:\mathrm{The}\:\mathrm{number}\:\mathrm{n}\:\mathrm{and} \\ $$$$\mathrm{n}+\mathrm{192}\:\mathrm{are}\:\mathrm{three}−\mathrm{digit}\:\mathrm{and} \\ $$$$\mathrm{four}−\mathrm{digit}\:\mathrm{palindromes}\:\mathrm{respectively}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{m}?\: \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{mathematically}? \\ $$

Answered by floor(10²Eta[1]) last updated on 29/Aug/20

we just need to get a 4 digit palindrome  that when subtracted by 192 we get a   palindrome with 3 digits  the first 4 digits palindromes are:  1001  1111  1221  ....  notice that we can stop here because if we   subtract 192 of 1221 we dont get a   3 digit number.  now we just need to test those 2 cases:  1001−192=809. not a palindrome.  1111−192=919. is a palindrome  so n=919 and the sum of the digits  is 9+1+9=19

$$\mathrm{we}\:\mathrm{just}\:\mathrm{need}\:\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{4}\:\mathrm{digit}\:\mathrm{palindrome} \\ $$$$\mathrm{that}\:\mathrm{when}\:\mathrm{subtracted}\:\mathrm{by}\:\mathrm{192}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\: \\ $$$$\mathrm{palindrome}\:\mathrm{with}\:\mathrm{3}\:\mathrm{digits} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{4}\:\mathrm{digits}\:\mathrm{palindromes}\:\mathrm{are}: \\ $$$$\mathrm{1001} \\ $$$$\mathrm{1111} \\ $$$$\mathrm{1221} \\ $$$$.... \\ $$$$\mathrm{notice}\:\mathrm{that}\:\mathrm{we}\:\mathrm{can}\:\mathrm{stop}\:\mathrm{here}\:\mathrm{because}\:\mathrm{if}\:\mathrm{we}\: \\ $$$$\mathrm{subtract}\:\mathrm{192}\:\mathrm{of}\:\mathrm{1221}\:\mathrm{we}\:\mathrm{dont}\:\mathrm{get}\:\mathrm{a}\: \\ $$$$\mathrm{3}\:\mathrm{digit}\:\mathrm{number}. \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{just}\:\mathrm{need}\:\mathrm{to}\:\mathrm{test}\:\mathrm{those}\:\mathrm{2}\:\mathrm{cases}: \\ $$$$\mathrm{1001}−\mathrm{192}=\mathrm{809}.\:\mathrm{not}\:\mathrm{a}\:\mathrm{palindrome}. \\ $$$$\mathrm{1111}−\mathrm{192}=\mathrm{919}.\:\mathrm{is}\:\mathrm{a}\:\mathrm{palindrome} \\ $$$$\mathrm{so}\:\mathrm{n}=\mathrm{919}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits} \\ $$$$\mathrm{is}\:\mathrm{9}+\mathrm{1}+\mathrm{9}=\mathrm{19} \\ $$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Thanks.

$$\mathrm{Thanks}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com