How many positive integers less than \(1000 \) have the property that the sum of the digits of each such number is divisible by \(7\) and the number itself is divisible by \(3\)?

No - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28

I believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.

I agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 432*1 = 24, not 4+3+2+1 = 10. What am I missing?

Hello,
There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3
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## Comments

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TopNewestLet's think of a number \(abc (0 \leq a, b, c \leq 9)\). \(a+b+c \equiv 0 (\mod 3 \text{and} \mod 7)\). Thus, \(a+b+c=21\).

\[(3, 9, 9) \rightarrow \frac{3!}{2}, (4, 8, 9) \rightarrow 3!, (5, 7, 9) \rightarrow 3!, (5, 8, 8) \rightarrow \frac{3!}{2}, (6, 6, 9) \rightarrow \frac{3!}{2}, (6, 7, 8) \rightarrow 3!, (7, 7, 7)\] \(3+6+6+3+3+6+1=28\)

Please tell me if there is any error.

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Nice method

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Good one brother

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What ans did you get?

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Hey Aaron. How much are u getting with bonus? With bonus i am getting 12.

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25 is the answer as per me

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No - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28

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28

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we just need to find the numbers which add upto 21

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Answer is 28

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How mañy have you got right in PRMO - 17?

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@Md Zuhair Is the paper for 9,10,11 and 12 same?

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Yes Sir!

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unfortunately, i will get only 10 questions correct. I did very silly mistakes

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Sir😅 I am getting 10 along with bonus!

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Oh. U mean 8/28 u r getting?

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well, if it is bonus that means 2 questions marks are given extra.If it was been written question deleted then scores would be evaluated out of 28

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Geometry was quite tough and lengthy! Excluding bonus , I'm getting 8

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oh...btw,where do u live (i mean which region)

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Rajasthan..U?

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oh

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So , you are already selected..Great 👍

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How already selected?

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Coz as per cutoff(s) uploaded by Resonance , cutoff in Chandigarh is lower than others ( Rajasthan , Maharashtra , UP , etc) ... That's why!

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According to Resonance , cutoff in Chandigarh is just 4(questions)

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Ya. Thats ridiculous. WB region has always got a higher cutoff...

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i don't think it will be so low

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If that isnt, then wb will be higher and i will surely not qualify

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It's 11 in Rajasthan ! 😅😒

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## LetTheFateDecide !!Bye

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which class are u in toshit?

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@Shreyan Chakraborty .. How much?

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JANI NA BAJE HOYECHE

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ANSWER IS 28....HAS A BIJECTION WITH a+b+c=21 WHERE 0<a,b,c<=9........

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Are na na.... I am not telling that. How much are you getting?

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@Shreyan Chakraborty The Hundreds digit can't be 1 or 2..

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yeah hundreds digit cant be 1,2

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@Md Zuhair @Vilakshan Gupta I haven't attempted one of the bonus question. Will I still get marks for it?

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I think the question can be cancelled as how can a person attempt to decinal answers and i had attempted ine. So i dunno.

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Yup.I think so.

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@Vilakshan Gupta Ok!

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I believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.

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Exactly

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I agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 4

32*1 = 24, not 4+3+2+1 = 10. What am I missing?Log in to reply

Ah - that's where the 28 comes from - much more mathematical than my just listing and counting them

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Hello, There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 You can check out for more queries related to the JEE EXAMS from the following compilation

<a href="https://scoop.eduncle.com/jee-main-exam-date-notification">JEE MAIN Exam dates and Notification</a>

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Hey I'm getting 8/27 from jharkhand as per the new answer key of hbcse.will I qualify???

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Lets see....

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Hey , what does

discountedactually refer to?Log in to reply

Are the marks gonna be added to everyone's total or the questions will be cancelled (lowering the cutoff)?

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They will be added to totsl

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The questions will be cancelled

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@Pokhraj Harshal Ok! Then I'm also getting the same...

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How much? Without the question?

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Which region are u from??

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WB rgion

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And what about the other participants and their marks from your school. I mean the averages and the highest marks

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@Pokhraj Harshal Rajasthan region!

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@Md Zuhair 8

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O i see....

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this question came in this year PRMO answer is 28

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its sum is divisibli by 21 using this you can solve

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27

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28

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33

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zuhair tui ki amk jiggesh korchish??

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Accha.. nijer whatsapp number ta de... whatsapp e kotha bolchi

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