# Find the least number which when divided by 2, 3, 4, 5, 6 leaves a remainder of 1 but it is divided by 7 completely.

I came across a question which is as follows:

Find out the smallest number which leaves remainder of 1 when divided by 2, 3, 4, 5, 6 but divided by 7 completely.

What I did is given below step wise.

Step 1- Find out the LCM of 2, 3, 4, 5, 6 which is 60.

Step 2- Add 1 to 60 which is 61.

Step 3- Multiple 61 by 7 repeatedly till it fulfills the condition that remainder should be 1.

Step 4- I got the answer 146461 which seems to correct.

So now my question is:

1) Is this answer correct? If yes how to verify that this is smallest number which fulfills above condition?

2) I think this is not the best way to do this question. So Can anyone give a better way to solve this problem?

Thanks in advance

• 301?? You could also use a system of congruence equations. By the way, you don't want to multiply by 61...Think about why... Jan 24, 2015 at 3:44
• How did you reach at 301 by using congruence relation? Can you pls explain it bit further? Jan 24, 2015 at 3:51
• The answer is less than $420$. Jan 24, 2015 at 4:02
• Well, first and foremost you were on the right track with LCM of 2,3,4,5,6. Adding 1 yields a remainder of 1. So numbers that leave a remainder of 1 would be 60k+1. Jan 24, 2015 at 4:03
• @Eleven-Eleven This method works, because eventually $7^{k}\equiv 1\pmod{60}$, so starting with $61$ and repeatedly multiply by $7$ works. It would work just as well to start with $1$ and repeated multiplying by $7$, of course, or you could apply the Chinese Remainder Theorem, as inmy answer. Repeated multiplying will not find the smallest answer, of course. Jan 24, 2015 at 4:17

## 6 Answers

You definitely need $n\equiv 1\pmod {60}$ and $n\equiv 0\pmod 7$. So the trick is to apply the Chinese remainder theorem. Solve $60x+7y=1$ with $(x,y)=(2,-17)$.

Then $n\equiv 1\cdot 7\cdot (-17)+0\cdot 60\cdot 2\pmod{420}$, or $n\equiv -119\pmod{420}$. The smallest such positive number is $420-119=301$.

Consider $60k+1,$ $$60(7n)+1=420n+1$$ $$60(7n+1)+1=420n+61$$ $$60(7n+2)+1=420n+121$$ $$60(7n+3)+1=420n+181$$ $$60(7n+4)+1=420n+241$$ $$60(7n+5)+1=420n+301$$ $$60(7n+6)+1=420n+361.$$ Now by divisibility test for 7, among $1, 61, 121, 181, 241, 301, 361$, only $301$ is divisible by $7.$

Therefore any number of the form $420n+301$ satisfies the given requirements.

Here's another way to tackle it. You already figured out that you're looking for $60k+1$. When you multiply $60$ by $k$ you want a predecessor to a multiplication of $7$.

$60 \equiv 4 \pmod 7$ and $5 \times 4 = 20$ which is a predecessor to a multiplication of $7$.

So $k=5$.

• The OP didn't "already figure out that [he's] looking for $60k+1$". He thought he was looking for $61\cdot 7^k$. Jan 24, 2015 at 12:58

Since no-one has mentioned it in their answers so far, your step 3 is wrong, or at least ill-advised, in that you are coming across a lot of answers that do not meet your carefully-set-up satisfaction of the first five conditions. For example, $61\times 7 = 427$ does not meet the desired remainders for $4$ or $5$.

The problem is that you have abandoned the safety of $n\equiv 1 \bmod 60$. The way to retain this in a simple search is to add 60 repeatedly looking for divisibility by 7.

We can do a bit better than that, though. We can translate into smaller numbers to make life easier for ourselves. $60 \equiv 4 \bmod 7$ (and of course $61 \equiv 5 \bmod 7$) so we can ask: how many 4s do we need to add to 5 before the answer is divisible by 7?

Again we can slog through the possibilities but the multiples of 7 are easier to cope with. We add two 4s (13 - and maybe see that we've reached $6 \bmod 7$) and add another two 4s to reach $21 \equiv 7 \equiv 0 \bmod 7$ - adding four 4s altogether. So back on our Actual Problem, we need to add four 60s - $4\times 60=240$ to our original $61$ so $240+61=301$ for the smallest positive solution.

Note that $60$ is your interval between satisfying those first 5 conditions, but we could also have started our search at $1$ rathe than $61$, with (of course) the same eventual result.

Below are a few different approaches (besides the standard Extended Euclidean Algorithm).

$${\rm mod}\ 60\!:\ x\equiv 1\equiv 7n\!\iff\! n\equiv\dfrac{1}7\equiv \dfrac{-59}7\equiv \dfrac{-119}7\equiv -17\,$$

therefore $$\smash[t]{\,x = 7(\overbrace{-17\!+\!60k}^{\large n}) = -119+420k}$$

Alternatively $$\:\!$$ mod $$\,60\!:\ \color{#c00}{7^4\equiv 1}\,$$ (by true mod $$3,4,5)\,$$ so $$\smash[b]{\,\color{#c00}{7^{-1}\equiv 7^3}\equiv 7(\underbrace{-11}_{\Large 7^2})\equiv -17}$$

Alternatively  we can employ  Inverse Reciprocity

$$\qquad\!\! \dfrac{1}7\ {\rm mod}\ 60\ \equiv\ \dfrac{1-60\overbrace{\left(\color{#c00}{\dfrac{1}{60}}\ {\rm mod}\ 7\right)}^{\Large \color{#0a0}{\equiv\, 2}}}7\,\equiv\, \dfrac{-119}7 \,\equiv\, -17\$$

where we've used: $$\ \ {\rm mod}\ 7\!:\,\ \color{#c00}{\dfrac{1}{60}}\equiv \dfrac{8}4\color{#0a0}{\equiv 2}\,\$$ (or recurse on $$\,\dfrac{1}{60}\bmod 7 \,\equiv\, \dfrac{1}4 \bmod 7\,)$$

See here and its links for further methods and elaboration.

Remark  In the first method we found a numerator $$\,-119\equiv 1\pmod{60}$$ that's also divisible by $$7$$ by brute force, i.e. we tested $$\,1-60k\,$$ for $$\,k=1,2\ldots$$ But now we see that the solution $$\,\color{#0a0}{ k\equiv 2}\,$$ is simply $$\, k = \color{#c00}{\dfrac{1}{60}}\,\bmod\, 7,\,$$ a "reciprocal" of our sought $$\,\dfrac{1}7\bmod 60\,$$ (i.e. swap $$\,7, 60)$$.

Beware $$\$$ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

• See also this answer to a similar congruence system, which contains further details needed for a complete rigorous proof. Mar 16, 2021 at 8:32

From $x\equiv 1 \pmod{60}$ we can proceed by adding the modulus until we find something congruent to $0\pmod{7}$:

$\pmod{60}:x\equiv 1 \equiv 61 \equiv 121 \equiv 181 \equiv 241 \equiv 301$.

Since $301\equiv 0\pmod{7}$, it gives the minimal solution.