Find-the-sum-of-all-three-digit-numbers-started-with-odd-number-when-each-digit-are-different-Please-help-

Question Number 188879 by Nimnim111118 last updated on 08/Mar/23

$$Findthesumofallthreedigitnumbers$$$$startedwithoddnumberwheneachdigit$$$$aredifferent.$$$$$$$$Pleasehelp\dots$$

Commented by mr W last updated on 08/Mar/23

$$doyouknowtheanswer?$$

Commented by Nimnim111118 last updated on 08/Mar/23

$$Sorry,Idontknowtheanswer.$$$$infact,Ihavenoidea$$

Commented by mr W last updated on 09/Mar/23

$$ihopemysolutionbelowcouldhave$$$$helpedyou.$$

Answered by mr W last updated on 09/Mar/23

$$axy$$$$xay$$$$xya$$$$[100\times 9\times 8+10\times 8\times 8+8\times 8](1+2+3+\dots +9)$$$$=355680✓$$

Commented by Nimnim111118 last updated on 08/Mar/23

$$Thankyousomuch.$$$$ButstillIdontgettheidea\dots$$

Commented by Nimnim111118 last updated on 08/Mar/23

$$Inmyopinion:$$$$Required:\bullet sumofall3digit$$$$\bullet allnumberstartedbyodd$$$$\bullet nodigitarerepeated$$$$So,ourquestionislikekytobeaddedup$$$$thefolowing$$$$102+103+104+105+106+107+108+109$$$$+120+123+124+125+126+127+128+129$$$$+130+132+134+135+136+137+138+139$$$$anditwillgoupto$$$$980+981+982+983+984+985+986+987$$$$$$

Commented by mr W last updated on 09/Mar/23

$$inthiswayyousumupallthevalid$$$$numbers.$$$$but{i}^{\prime }mtoolazytodothishardwork.$$$$insteadofthisisumupallthedigits$$$$inthenumbers.thisiseasier,because$$$$wehavemanymanynumbers,but$$$$only9digits.$$

Commented by mr W last updated on 08/Mar/23

$$inmysolutionaboveiconsidered$$$$all3digitnumberswithdifferent$$$$digits,alsonumbersstartingwithan$$$$evendigit,like235.ifyoumeanonly$$$$numbersstartingwithanodddigit,$$$$thenimustmodifymysolution.$$

Answered by mr W last updated on 09/Mar/23

$$whenwewanttogetthesumofsome$$$$numbers,e.g.123+456,wecanalsoget$$$$theresultbysummingupthevalues$$$$whichtheindividualdigitsrepresent.$$$$inthisexample,toget123+456,we$$$$canproceedlikethis:$$$$100\times (1+4)+10\times (2+5)+1\times (3+6)$$$$=500+70+9$$$$=579$$$$tosumupall3digitnumberswhich$$$$startwithanodddigitandconsist$$$$of3differentdigits,wecanusethis$$$$method.$$$$togetthesumofallvalueswhich$$$$thedigit“a“representsinall$$$$such3digitnumbers:$$$$\underset{―}{typeaxy:}$$$$a=1,3,5,7,9$$$$foreach“a^{″}thereare9\times 8numbers,$$$$100\times 9\times 8\times (1+3+5+7+9)=180000$$$$\underset{―}{typexay:}$$$$a=1,3,5,7,9or2,4,6,8$$$$x=1,3,5,7,9$$$$10\times 4\times 8\times (1+3+5+7+9)$$$$+10\times 5\times 8\times (2+4+6+8)=16000$$$$\underset{―}{typexya:}$$$$similarlytotypexay.$$$$1\times 4\times 8\times (1+3+5+7+9)$$$$+1\times 5\times 8\times (2+4+6+8)=1600$$$$\mathit{a}\mathit{l}\mathit{l}\mathit{t}\mathit{o}\mathit{g}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{r}:$$$$180000+16000+1600=197600✓$$$$$$$$thatmeanstherearetotally$$$$5\times 9\times 8=360suchnumbers,andthe$$$$sumofthemis197600.$$

Commented by mr W last updated on 08/Mar/23

$$maybeBrandonsircancheckthis$$$$resultwithaprogram.$$

Commented by ARUNG_Brandon_MBU last updated on 09/Mar/23

Commented by ARUNG_Brandon_MBU last updated on 09/Mar/23

#include <iostream> using namespace std; int main(void) { short count=0; int sum=0; for (short i0, i1, i2, i=102; i<1000; i++) { i0 = i / 100; i1 = i % 100 / 10; i2 = i % 10; if (i0%2==1&&i0!=i1&&i0!=i2&&i1!=i2) { sum += i; count++; } } cout << "There are " << count << "numbers"; cout << "\n their sum is " << sum; return 0; }

Commented by ARUNG_Brandon_MBU last updated on 09/Mar/23

Commented by mr W last updated on 09/Mar/23

$$greatprograming!thanksalotsir!$$

Commented by ARUNG_Brandon_MBU last updated on 09/Mar/23

You're welcome, Sir ����

Terms of Service
Privacy Policy
Contact: info@tinkutara.com