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!

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TopNewestFill the thousands, hundreds, and tens digits arbitrarily. Now there are exactly three ways to fill the ten-thousands digit. This easily gives \(10^3 \cdot 3 = \boxed{3000}\) numbers.

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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.

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Nice solution!

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Thanx bud!

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Damn simple prob!Two second question!

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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

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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 = 3000Log in to reply

30

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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.

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Please help me in my approach:

Let the required \(5\) digit number be \(\overline{a_1a_2a_3a_44}\)

Since the number is already divisible by \(2\) we need to assure the divisibility of \(3\)

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

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

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

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No need to complicate it this much bud!

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

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

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