Multiplying and Dividing Integers by Powers of 10

\(34 \times 100 = 3400\). Meanwhile, \(34 \times 10 = 340\), then \(340 \times 10 = 3400\). Both give 3400. This works because \(100 = 10 \times 10\), so multiplying by 100 is the same as multiplying by 10 twice.

In terms of place value, multiplying by 100 shifts every digit two places to the left. Multiplying by 10 twice shifts every digit one place to the left each time — which is the same thing. The 3 moves from tens to thousands, the 4 from ones to hundreds, and two zeros fill the empty columns.

\(5 \div 1000 = 0.005\). Dividing by 1000 shifts every digit three places to the right. The 5 moves from the ones column to the thousandths column. Since 5 is much smaller than 1000, the result must be less than 1.

Think of it as sharing: if you split 5 into 1000 equal parts, each part is tiny — just five thousandths. This challenges the idea that dividing a whole number always gives another whole number.

Work through it step by step: \(4000 \div 10 = 400\), then \(400 \div 10 = 40\), then \(40 \div 10 = 4\). Each division shifts every digit one place to the right. Three successive divisions shift the digits three places right in total.

This is the same as dividing by \(10 \times 10 \times 10 = 1000\) in one go: \(4000 \div 1000 = 4\). Just as multiplying by 100 is the same as multiplying by 10 twice, dividing by 1000 is the same as dividing by 10 three times. The exponent in the power of 10 tells you how many one-place shifts to make.

Look at the pattern: \(10^3 = 1000\), \(10^2 = 100\), \(10^1 = 10\). Each time the power decreases by 1, we divide by 10. Following this pattern: \(10^0 = 10 \div 10 = 1\).

Alternatively, \(10^1 \div 10^1 = 10^{1-1} = 10^0\). Since any number divided by itself equals 1, \(10^0\) must be 1. Students often assume the zero exponent means “nothing” and so guess 0 — but the exponent tells you how many 10s to multiply, and multiplying by no 10s at all just leaves you with 1.

Because of the commutative property of multiplication, the order of the numbers doesn’t matter. Just as \(3 \times 4\) is the same as \(4 \times 3\), \(100 \times 45\) will naturally yield the same product as \(45 \times 100\).

Students often view “\(\times 100\)” merely as an action or rule applied to a starting number, failing to see the relationship goes both ways. It is usually easier to think about shifting the digits of 45 than doing standard multiplication for “100 multiplied by 45”, but mathematically they are identical operations.

Example: 3 (3 × 100 = 300)

Another: 21 (21 × 100 = 2100)

Creative: −7 (−7 × 100 = −700) — negative integers work too. Or 99 (99 × 100 = 9900).

Trap: 50 (50 × 100 = 5000, which ends in three zeros, not two). If the integer already ends in a zero, multiplying by 100 adds two more, creating more trailing zeros than expected. The integer must not end in zero itself.

Example: 15 (15 × 10 = 150)

Another: 50 (50 × 10 = 500)

Creative: 10 (10 × 10 = 100) — right on the lower boundary, the smallest integer that works. Or 99 (99 × 10 = 990), right on the upper boundary.

Trap: 100 (100 × 10 = 1000, which has four digits, not three). A student might think any “small” number works, but once the integer reaches 100, multiplying by 10 pushes the result past 999 into four-digit territory.

Example: 70 ÷ 10 = 7

Another: 500 ÷ 100 = 5

Creative: 8,000,000 ÷ 1,000,000 = 8 — using an enormous power of 10. Or 0 ÷ 10 = 0 — zero is technically a single-digit result!

Trap: 100 ÷ 10 = 10 — this gives a two-digit number, not a single digit. A student might think “just remove a zero from 100” and not notice that the result has two digits.

Example: −1 (−1 × 100 = −100, and −100 < −1)

Another: −5 (−5 × 100 = −500, and −500 < −5)

Creative: −1000 (−1000 × 100 = −100,000) — a large negative becomes enormous in the negative direction.

Trap: 2 (2 × 100 = 200, and 200 > 2, not less). Any positive integer gets larger when multiplied by 100. The key insight is that only negative integers satisfy this condition: multiplying by 100 moves them further from zero in the negative direction.

True for positive integers: \(5 \times 10 = 50\), and 50 > 5. But false for negative integers: \(-3 \times 10 = -30\), and −30 is less than −3 (it’s further from zero in the negative direction).

And for zero: \(0 \times 10 = 0\), so the number doesn’t change at all. The misconception “multiplying always makes things bigger” only holds for positive numbers multiplied by a value greater than 1.

True when the integer is a multiple of that power of 10: \(300 \div 10 = 30\) and \(5000 \div 100 = 50\). But false otherwise: \(7 \div 10 = 0.7\) and \(23 \div 100 = 0.23\) — neither is a whole number.

This targets the misconception that dividing integers always gives integers. In fact, dividing by a power of 10 only gives a whole number when the integer has enough trailing zeros to “absorb” the division.

The non-zero digits in the answer are always exactly the same as the digits in the original number. For example, \(34 \times 100 = 3400\) — the 3 and 4 are still there, just in different positions. The only “new” digits are placeholder zeros filling the empty columns.

This is the core principle: multiplying by a power of 10 shifts digits to new positions but never changes or creates non-zero digits. Recognising this means students don’t need to perform a standard multiplication — they just shift and fill with zeros.

This is always true. For example: \(7 \times 100 = 700\), and \(700 \div 100 = 7\). It works for negatives: \(-4 \times 10 = -40\), and \(-40 \div 10 = -4\). And for zero: \(0 \times 1000 = 0\), and \(0 \div 1000 = 0\).

Multiplying and dividing by the same power of 10 are inverse operations — each one undoes the other. This is a useful checking strategy: if you calculate \(36 \times 100 = 3600\), you can verify by checking \(3600 \div 100 = 36\). If the digits shift left when multiplying, they shift right by the same amount when dividing.

It is true for positive integers: \(500 \div 10 = 50\), and 50 is less than 500. However, it is false for negative integers: \(-50 \div 10 = -5\). Since −5 is closer to zero on a number line than −50, it is actually a larger number!

It is also false for zero: \(0 \div 10 = 0\), which is exactly the same size. This directly challenges the deeply held assumption that “division always makes things smaller.”

The student shifted the digits only one place to the left instead of two. This is the “miscounting the place-value shift” error. \(45 \times 10 = 450\), but \(45 \times 100 = 4500\).

Since \(100 = 10^2\), multiplying by 100 requires shifting every digit two places to the left. The 4 moves from tens to thousands, the 5 from ones to hundreds, and two zeros fill the empty columns. A good check: the answer should be roughly 100 times bigger than the starting number.

The answer is correct — 60 ÷ 10 does equal 6 — but the reasoning is dangerously wrong. The student is using the “remove a zero” trick, which is a superficial shortcut rather than genuine understanding.

This rule breaks down immediately when the number doesn’t end in zero: what is \(63 \div 10\)? The student’s method offers no way forward, yet the answer is simply 6.3 — each digit shifts one place to the right. The correct reasoning is that dividing by 10 moves every digit one place-value column to the right, whether or not the number ends in zero.

This trap is closely related to the “move the decimal point” misconception. The decimal point is an anchor separating whole numbers from fractions; it never physically moves. It is the digits that shift across the static place value columns.

The student has confused an exponent with multiplication. They read \(10^3\) as “\(10 \times 3 = 30\)” instead of “\(10 \times 10 \times 10 = 1000\).” This is the “exponent means multiply” misconception — treating the power as a multiplier rather than a repeated multiplication.

The correct calculation is \(8 \times 10^3 = 8 \times 1000 = 8000\). The exponent tells us how many times 10 is multiplied by itself: \(10^3 = 10 \times 10 \times 10 = 1000\), not \(10 \times 3 = 30\). The answer should be far larger than 240.

The student has the order exactly backwards. They applied the rule “multiplying by a bigger number gives a bigger result,” which is true for positive numbers but fails for negatives. When a negative number is multiplied by a larger positive value, it moves further from zero in the negative direction, making it smaller.

−5000 is the smallest (furthest left on a number line), then −500, then −50 is the largest (closest to zero). The correct order from smallest to largest is: −5000, −500, −50.