To find four numbers that multiply to 56, one possible combination is 1, 2, 2, and 14 (1 × 2 × 2 × 14 = 56). Another combination is 2, 2, 7, and 2 (2 × 2 × 7 × 2 = 56). There are multiple sets of numbers that can achieve this product, depending on the chosen factors.
As a digit in other numbers it appears 20 times
To form three-digit even numbers from the set {2, 3, 5, 6, 7}, we can use the digits 2 or 6 as the last digit (to ensure the number is even). For each case, we can choose the first two digits from the remaining four digits. For three-digit numbers, there are 2 options for the last digit and (4 \times 3 = 12) combinations for the first two digits, resulting in (2 \times 12 = 24) even numbers. For four-digit even numbers, we again have 2 options for the last digit. The first three digits can be selected from the remaining four digits, giving us (4 \times 3 \times 2 = 24) combinations for each last digit. Thus, there are (2 \times 24 = 48) even four-digit numbers. In total, there are (24 + 48 = 72) three-digit and four-digit even numbers that can be formed from the set.
Twelve of them.
No.No.No.No.
When a person writes all 4-digit numbers, they range from 1000 to 9999, totaling 9000 numbers. The digit '2' can appear in each of the four positions (thousands, hundreds, tens, and units). In each position, '2' will appear 900 times (e.g., for the thousands place, numbers from 2000 to 2999), leading to a total of 3600 occurrences of the digit '2' across all four positions.
Yes.
As a digit in other numbers it appears 20 times
13.
To form three-digit even numbers from the set {2, 3, 5, 6, 7}, we can use the digits 2 or 6 as the last digit (to ensure the number is even). For each case, we can choose the first two digits from the remaining four digits. For three-digit numbers, there are 2 options for the last digit and (4 \times 3 = 12) combinations for the first two digits, resulting in (2 \times 12 = 24) even numbers. For four-digit even numbers, we again have 2 options for the last digit. The first three digits can be selected from the remaining four digits, giving us (4 \times 3 \times 2 = 24) combinations for each last digit. Thus, there are (2 \times 24 = 48) even four-digit numbers. In total, there are (24 + 48 = 72) three-digit and four-digit even numbers that can be formed from the set.
4*6+1
This will be difficult to answer accurately without knowing each set of numbers.
Twelve of them.
No.No.No.No.
The teams have played each other in four competitive games. Dundee United have won all four.
24, 25, 26, 27 (next to each other and add up to 102)
2 x 3 x 5 x 5 = 150
5x5x5x3=375