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 $3$?

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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

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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Your method is quite efficient vis-a-vis mine. The latter involved manual trials with 3 digit integers with integer 1 to 9 at the unit. Thank you

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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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