Let's see who really can count numbers

How many 5 digit positive integers are there such that the unit's place is 4 and it is divisible by 6?

Nice proofs and solutions are always welcome!

Note by Nihar Mahajan
4 years, 6 months ago

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Fill the thousands, hundreds, and tens digits arbitrarily. Now there are exactly three ways to fill the ten-thousands digit. This easily gives 1033=300010^3 \cdot 3 = \boxed{3000} numbers.

Ivan Koswara - 4 years, 6 months ago

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There are 15000 5 digit numbers that are divisible by 6.The first i.e the smallest one ending in 2.The unit digit of these numbers make a pattern,2,8,4,0,6,2,8,4,0,6,2....Hence 1 in every 5 consecutive numbers ends in 4.Hence answer to the problem is ,15,000/5=3000.

Adarsh Kumar - 4 years, 6 months ago

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Damn simple prob!Two second question!

Adarsh Kumar - 4 years, 6 months ago

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Nice solution!

Kalash Verma - 4 years, 6 months ago

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Thanx bud!

Adarsh Kumar - 4 years, 6 months ago

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Please help me in my approach:

Let the required 55 digit number be a1a2a3a44\overline{a_1a_2a_3a_44}

Since the number is already divisible by 22 we need to assure the divisibility of 33

So , i=14ai42(mod3)\displaystyle\sum_{i = 1}^4 a_i \equiv -4 \equiv 2 \pmod{3}

So , i=14ai=3k+2\displaystyle\sum_{i =1}^4 a_i = 3k+2 where {1k111a190a2,a3,a49\begin{cases} 1 \leq k \leq 11 \\ 1 \leq a_1 \leq 9 \\ 0 \leq a_2,a_3,a_4 \leq 9 \end{cases}

Indeed , finding solutions to this equation will fetch the required answer.

Nihar Mahajan - 4 years, 6 months ago

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No need to complicate it this much bud!

Adarsh Kumar - 4 years, 6 months ago

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But we must learn different methods.May be this method can help in some other problem.

Nihar Mahajan - 4 years, 6 months ago

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@Nihar Mahajan But if there are two different methods to solve a prob we must take the shorter solution.Donn'y u think?

Adarsh Kumar - 4 years, 6 months ago

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@Adarsh Kumar I know we must have the shorter one , but if the shorter method is not applicable to some other problem then what will you do? If we do not get a pattern in some other problem this method will not work.

Nihar Mahajan - 4 years, 6 months ago

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@Nihar Mahajan True!Let me tell you the method of solving this.Because ai9a_i\leq 9,you can assume a1=9x,a2=9y,a3=9z,a4=9pa_1=9-x,a_2=9-y,a_3=9-z,a_4=9-p.Notice here that x,y,z,px,y,z,p are non-negative integers.Now you can consider every case and use stars and bars.But it would be very tedious.

Adarsh Kumar - 4 years, 6 months ago

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@Adarsh Kumar Yeah , thats a great idea. Let me see what happens next.

Nihar Mahajan - 4 years, 6 months ago

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@Nihar Mahajan Learn mthods that increase problem solving speed and increase accuracy, complicating things may take you down. :)

Parth Lohomi - 4 years, 6 months ago

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@Parth Lohomi Of course , a short and quick method is always preferred.But we must know different approaches to solve.May be the "short and quick" method does not work in some problems.In that case such different approaches may prove useful.Hence I prefer doing the same problem using different methods , rather that leaving it by solving it with the shortest method. :)

Nihar Mahajan - 4 years, 6 months ago

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It can also be seen as the number of four digit numbers that are divisible by 3, because the rest of the digits simply need to add upto a number that is one less than a multiple of 3. Under 1000 you have 333 such numbers and under a 10000 you have 3333 numbers divisible by 3. So we get 3000 by subtraction.

vishnu c - 4 years, 6 months ago

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30

Shayaan Subzwari - 4 years, 6 months ago

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First term is 10014 and last term is 99984 (can easily be found by hit and trial method). These are the corresponding terms of an arithmetic progression with common difference = 30. An = a + (n-1)d, 99984 = 110014 + (n-1) 30, n = 3000

Rahul Mehta - 4 years, 6 months ago

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The numbers are of the form 10k+4. We set 10k+4=6z or 5k+2=3z. We see that when k=-1 and z = -1 there is a solution note that when k= -1+3b and z = -1+5b. 5k+2= 5(-1+3b)+2=15b-3=3(5b-1)=3z. By euclidean algorithm there is only one solution in the interval where k= -2,-1,0 and z= -4,-3,-2,-1,0. eg one solution for (k,w) in this specified domain. Now we have 6z=6(-1+5b)=30b-6. All such numbers are of this form. Now 10000=<30b-6<=99999. 334<=b<=3333. Number of integers in the range = no of integers from 1 to 3333 - no of integers from 1 to 333 = 3333-333=3000

Siddharth Iyer - 4 years, 5 months ago

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we form a AP here : 10014, 10044, 10074............................................................99984 therefore no. of terms n= 99984-10014/30 +1 =3000

Aakash Khandelwal - 4 years, 6 months ago

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