School curriculum· 20 min· 22 questions
Natural numbers are the foundation of mathematics. They rarely appear as a separate exam topic, yet almost every problem needs the order of operations, division with remainder or a search through natural solutions. This lesson runs from place value to "find the least value" problems.
Natural numbers are the counting numbers 1, 2, 3, …; the set is written N. Zero is not a natural number — a detail exams like to test.
Two-digit number: ab = 10a + b; three-digit: abc = 100a + 10b + c (a ≠ 0)
This expansion is the key to digit problems. For example, a two-digit number minus its reversal is always divisible by 9: (10a + b) − (10b + a) = 9(a − b).
Number of n-digit numbers: 9·10ⁿ⁻¹ (90 two-digit, 900 three-digit)
Sums and products of natural numbers are always natural; differences and quotients need not be: 3 − 5, 7 : 2.
Commutative: a + b = b + a, a·b = b·a
Associative: (a + b) + c = a + (b + c), (ab)c = a(bc)
Distributive: a·(b + c) = a·b + a·c
The distributive law speeds up mental arithmetic: 37·99 = 37·(100 − 1) = 3700 − 37 = 3663.
Order of operations: 1) brackets; 2) powers; 3) multiplication and division (left to right); 4) addition and subtraction (left to right).
Evaluate 48 : 4·3 − 2·(5 − 3)².
Note: 48 : 4·3 is not 48 : 12 = 4 — work left to right.
For any a and b > 0 there are unique q (quotient) and r (remainder):
a = b·q + r, where 0 ≤ r < b
Remainders of sums and products come from the remainders of the parts. For 23·17 divided by 5: 23 → 3, 17 → 2, 3·2 = 6 → 1, so the remainder is 1.
A number divided by 7 gives quotient 15 and remainder 4. Find it.
Problems about natural numbers are often phrased with sets.
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
U = {x | −10 ≤ x ≤ 10, x ∈ Z}, A = {−7, …, 3}, B = {−3, …, 7}. How many elements are in the complement of A ∪ B?
For "find the least value for natural numbers" problems, listing every possible pair in order is the safest method.
If a·b = 30, find the least value of a + 2b − 1.
Reasoning: b counts double, so look for a small b without making a huge.
In the decimal system the value of each digit depends on its place (§1). So in problems about digits the main technique is to write the number as a sum of place-value terms.
A two-digit number: ab = 10a + b, where a ∈ {1, …, 9}, b ∈ {0, …, 9}
A three-digit number: abc = 100a + 10b + c
Several useful facts follow from this notation:
Another well-known technique is the sum of consecutive natural numbers. According to the story, the young Gauss found 1 + 2 + … + 100 by pairing the outer numbers: 1 + 100 = 2 + 99 = … = 101; there are 50 such pairs, so the sum is 5050. In general:
1 + 2 + … + n = n(n + 1)/2
In a digit problem always check the answer against the conditions: digits must lie between 0 and 9, and the leading digit cannot be 0.
The digits of a two-digit number add up to 9. If the digits are swapped, the number increases by 27. Find the number.