# How to do Long Division Step by Step for Kid?

Long division is a versatile method for handling complex divisions without using a calculator. It is the preferred method when dividing by a number with two or more digits, particularly if the division is not exact. It can be used to calculate a remainder or give an answer to a paticular number of decimal places.

In order to demonstrate the method we'll work through the solution to 2738 ÷ 70.

### Step 1

Long division works from left to right. Since 70 is a 2-digit number, it will not go into 2, the first digit of 2738, and so successive digits are added until a number greater than 70 is found. In this case 2 digits are added to make 273. Note the other digits in the original number have been turned grey to emphasise this and grey zeroes have been placed above to show where division was not possible with fewer digits.
The closest we can get to 273 without exceeding it is 210 which is 3 × 70. These values have been added to the division, highlighted in red.
 0 0 3 70 2 7 3 8 2 1 0
You will notice that the division is set out carefully with the digits in vertical columns. This is very important when you work them out by hand.

### Step 2

Next, work out the remainder by subtracting 210 from 273. This gives us 63. Bring down the 8 to make a new target of 638.
 3 70 2 7 3 8 2 1 0 6 3 8
The digit brought down and the new target have been highlighted in blue.

### Step 3

With a target of 638, the closest we can get is 630 by multiplying 70 by 9. Write 9 in the next column of the answer, and 630 below the 638 as shown.
 3 9 70 2 7 3 8 2 1 0 6 3 8 6 3 0

### Step 4

Finally, subtract 630 from 638 giving 8. Since there are no other digits to bring down, 8 is therefore also the remainder for the whole sum.
So 2738 ÷ 70 = 39 rem 8
 3 9 70 2 7 3 8 2 1 0 6 3 8 6 3 0 8
Solution: 2738 ÷ 70 = 39 r 8

### Step 1

Long division works from left to right. Since 47 is a 2-digit number, it will not go into 4, the first digit of 46423, and so successive digits are added until a number greater than 47 is found. In this case 2 digits are added to make 464. Note the other digits in the original number have been turned grey to emphasise this and grey zeroes have been placed above to show where division was not possible with fewer digits.
The closest we can get to 464 without exceeding it is 423 which is 9 × 47. These values have been added to the division, highlighted in red.
 0 0 9 rem 34 47 4 6 4 2 3 4 2 3

 47 × table 1 × 47 = 47 2 × 47 = 94 3 × 47 = 141 4 × 47 = 188 5 × 47 = 235 6 × 47 = 282 7 × 47 = 329 8 × 47 = 376 9 × 47 = 423

### Step 2

Next, work out the remainder by subtracting 423 from 464. This gives us 41. Bring down the 2 to make a new target of 412.
 9 rem 34 47 4 6 4 2 3 4 2 3 4 1 2

 47 × table 1 × 47 = 47 2 × 47 = 94 3 × 47 = 141 4 × 47 = 188 5 × 47 = 235 6 × 47 = 282 7 × 47 = 329 8 × 47 = 376 9 × 47 = 423

### Step 3

With a target of 412, the closest we can get is 376 by multiplying 47 by 8. Write 8 in the next column of the answer, and 376 below the 412 as shown.
 9 8 rem 34 47 4 6 4 2 3 4 2 3 4 1 2 3 7 6

 47 × table 1 × 47 = 47 2 × 47 = 94 3 × 47 = 141 4 × 47 = 188 5 × 47 = 235 6 × 47 = 282 7 × 47 = 329 8 × 47 = 376 9 × 47 = 423

### Step 4

Next, work out the remainder by subtracting 376 from 412. This gives us 36. Bring down the 3 to make a new target of 363.
 9 8 rem 34 47 4 6 4 2 3 4 2 3 4 1 2 3 7 6 3 6 3

 47 × table 1 × 47 = 47 2 × 47 = 94 3 × 47 = 141 4 × 47 = 188 5 × 47 = 235 6 × 47 = 282 7 × 47 = 329 8 × 47 = 376 9 × 47 = 423

### Step 5

With a target of 363, the closest we can get is 329 by multiplying 47 by 7. Write 7 in the next column of the answer, and 329 below the 363 as shown.
 9 8 7 rem 34 47 4 6 4 2 3 4 2 3 4 1 2 3 7 6 3 6 3 3 2 9

 47 × table 1 × 47 = 47 2 × 47 = 94 3 × 47 = 141 4 × 47 = 188 5 × 47 = 235 6 × 47 = 282 7 × 47 = 329 8 × 47 = 376 9 × 47 = 423

### Step 6

Finally, subtract 329 from 363 giving 34. Since there are no other digits to bring down, 34 is therefore also the remainder for the whole sum.
So 46423 ÷ 47 = 987 rem 34
 9 8 7 rem 34 47 4 6 4 2 3 4 2 3 4 1 2 3 7 6 3 6 3 3 2 9 3 4

 47 × table 1 × 47 = 47 2 × 47 = 94 3 × 47 = 141 4 × 47 = 188 5 × 47 = 235 6 × 47 = 282 7 × 47 = 329 8 × 47 = 376 9 × 47 = 423

### 2nd Vietnamese Secondary Mathematics Contest

A2. The points X, Y and Z are chosen on the side BC, CA, AB of a given triangle ABC, respectively such that BX = CY = AZ. Prove that the triangle XYZ is equilateral if and only if so is the triangle ABC.

A3. Give a collection of real numbers A = (a1, a2,...an ), denote by A(2) the 2-sums set of A, which is the set of all sum a1+ai, for 1 ≤ i < j ≤ n. Give that A(2) = (2,2,3,3,4,4,4,4,4,5,5,5,6,6), determine the sum of squares of all elements of the original set A.

A4. Let M be a point on a given line segment AB. Draw three semicircles whose diameters are AM, BM, AB respectively, such that they are on the same side with respect to AB. Let I be the incenter and r be the inradius of the curvilinear triagle ABM (whose sides are the three semicircles just contructed). Prove that when M moves on the line segment AB, the locus of I is an are of an ellipse whose spanning chord passes through one of its foci.

### 1st Vietnamese Secondary Mathematics Contest

A2. Let n be a positive integer and let U(n) = {d1; d2;...dm} be the set of all positive divisors of n. Prove that
d12 + d22 + ...+ dm2 <

A3. Prove that

where a, b, c are there positive numbers satisfying abc = 1.

A4. Solve the equation

A5.  Let ABCD be a square, M is a point lying on CD (MC, MD). Through the point C draw
a line perpendicular to AM at H, BH meets AC at K. Prove that:

1) MK is always parallel to a fixed line when M moves on the side CD.

2) The circumcenter of the quadrilateral ADMK lies on a fixed line.

A6. Let a, b, c be positive real numbers such that abc = 1. Prove that

(For upper secondary schools)

A7. Conside all triangles ABC where A < B < C≤
Find the least value of the expression: M = cot2A + cot2B + cot2C + 2(cotA - cotB)(cotB - cotC)(cotC - cotA).

A8. Suppose that the tetrahedron ABCD statisfies the following conditions:  All faces are acute triangles and BC is perpendicular to AD.  Let ha, hd be respectively the lengths of the altitudes from A, D onto the opposite faces, and let 2α  be the measure of the dihedral angle at edge BC, d is the distance between BC and AD. Prove the inequalityl: