How about 4 times 7 is 28 and 8 divded by 2 is 4
Four of them: 4 times 7 = 28
To use four sevens to make 490, you can express it as follows: ( (7 \times 7 \times 10) - (7 + 7) ). Here, ( 7 \times 7 = 49 ), and multiplying by 10 gives you 490. The subtraction of ( 7 + 7 ) is not necessary for this specific equation, but you can also think of alternative ways to use four sevens and additional operations to reach the same result.
99 divided by 0.99
In canasta, a meld of sevens can be made using wild cards, such as jokers or twos. However, when forming a canasta, at least one natural seven (without wild cards) must be included in the meld. Additionally, the meld must consist of a minimum of four cards to be considered a valid canasta. Therefore, while you can include wild cards, you can't create a canasta of sevens entirely out of wild cards.
As a solution to the four fours problem (using the number four no more than four times to come up with the solution, 31 is equal to (4!+4)/4+4!
((7+7)*7) - 7
Four of them: 4 times 7 = 28
Well, isn't that a happy little problem to solve! To make 100 using four sevens, you can try this: 77 + 7/7 = 100. Just like painting a beautiful landscape, sometimes it's about mixing and blending things together in a creative way to achieve your desired result.
You could go (7+7+7)7=147
To use four sevens to make 490, you can express it as follows: ( (7 \times 7 \times 10) - (7 + 7) ). Here, ( 7 \times 7 = 49 ), and multiplying by 10 gives you 490. The subtraction of ( 7 + 7 ) is not necessary for this specific equation, but you can also think of alternative ways to use four sevens and additional operations to reach the same result.
99 divided by 0.99
eleven to the power of eleven
5*4+3+2
In canasta, a meld of sevens can be made using wild cards, such as jokers or twos. However, when forming a canasta, at least one natural seven (without wild cards) must be included in the meld. Additionally, the meld must consist of a minimum of four cards to be considered a valid canasta. Therefore, while you can include wild cards, you can't create a canasta of sevens entirely out of wild cards.
As a solution to the four fours problem (using the number four no more than four times to come up with the solution, 31 is equal to (4!+4)/4+4!
(8x8)=64
4! - 4 + 4/4