Use the Lattice method to solve the following:

Write 675 across the top and 58 down the side.

• $5 \times 6 = 30$
• $5 \times 7 = 35$
• $5 \times 5 = 25$
• $8 \times 6 = 48$
• $8 \times 7 = 56$
• $8 \times 5 = 40$

• Units diagonal: $0 = 0$
• Tens diagonal: $5 + 4 + 6 = 15$ (write 5, carry 1)
• Hundreds diagonal: $2 + 5 + 8 + 5 + 1 (\text{carry}) = 21$ (write 1, carry 2)
• Thousands diagonal: $3 + 4 + 2 (\text{carry}) = 9$
• Ten Thousands diagonal: $3 = 3$

The result is 39150.

b) $974 \times 79$

Write 974 across the top and 79 down the side.

• $7 \times 9 = 63$
• $7 \times 7 = 49$
• $7 \times 4 = 28$
• $9 \times 9 = 81$
• $9 \times 7 = 63$
• $9 \times 4 = 36$

• Units diagonal: $6 = 6$
• Tens diagonal: $8 + 3 + 3 = 14$ (write 4, carry 1)
• Hundreds diagonal: $2 + 9 + 1 + 6 + 1 (\text{carry}) = 19$ (write 9, carry 1)
• Thousands diagonal: $3 + 4 + 8 + 1 (\text{carry}) = 16$ (write 6, carry 1)
• Ten Thousands diagonal: $6 + 1 (\text{carry}) = 7$

The result is 76946.

c) $795 \times 74$

Write 795 across the top and 74 down the side.

• $7 \times 7 = 49$
• $7 \times 9 = 63$
• $7 \times 5 = 35$
• $4 \times 7 = 28$
• $4 \times 9 = 36$
• $4 \times 5 = 20$

• Units diagonal: $0 = 0$
• Tens diagonal: $5 + 6 + 2 = 13$ (write 3, carry 1)
• Hundreds diagonal: $3 + 3 + 8 + 6 + 1 (\text{carry}) = 21$ (write 1, carry 2)
• Thousands diagonal: $9 + 2 + 2 (\text{carry}) = 13$ (write 3, carry 1)
• Ten Thousands diagonal: $4 + 1 (\text{carry}) = 5$

The result is 58930.

2. Circle the number that is divisible by [missing divisor]

The divisor is missing from the question. Assuming the question asks to circle the numbers divisible by 6 (a common divisibility rule for elementary math), we check for divisibility by both 2 and 3.

Numbers: $42, 24, 107, 318, 195, 23, 84$

• A number is divisible by 2 if its last digit is even.
• A number is divisible by 3 if the sum of its digits is divisible by 3.
• A number is divisible by 6 if it is divisible by both 2 and 3.

• 42 (even)
• 24 (even)
• 107 (odd)
• 318 (even)
• 195 (odd)
• 23 (odd)
• 84 (even)

• $42 \implies 4+2=6$ (divisible by 3)
• $24 \implies 2+4=6$ (divisible by 3)
• $107 \implies 1+0+7=8$ (not divisible by 3)
• $318 \implies 3+1+8=12$ (divisible by 3)
• $195 \implies 1+9+5=15$ (divisible by 3)
• $23 \implies 2+3=5$ (not divisible by 3)
• $84 \implies 8+4=12$ (divisible by 3)

The numbers divisible by 6 are: $\boxed{42, 24, 318, 84}$.

3. Circle the number that is divisible by [missing divisor]

The divisor is missing from the question. Assuming the question asks to circle the numbers divisible by 6, we check for divisibility by both 2 and 3.

Numbers: $54, 210, 108, 144, 212, 96$

• 54 (even)
• 210 (even)
• 108 (even)
• 144 (even)
• 212 (even)
• 96 (even) All numbers in this list are divisible by 2.

• $54 \implies 5+4=9$ (divisible by 3)
• $210 \implies 2+1+0=3$ (divisible by 3)
• $108 \implies 1+0+8=9$ (divisible by 3)
• $144 \implies 1+4+4=9$ (divisible by 3)
• $212 \implies 2+1+2=5$ (not divisible by 3)
• $96 \implies 9+6=15$ (divisible by 3)

The numbers divisible by 6 are: $\boxed{54, 210, 108, 144, 96}$.

4. Solve the ff (following):

a) $6 \overline{)567}$

$56 \div 6 = 9$ with a remainder of 2.

Divide 27 by 6. $27 \div 6 = 4$ with a remainder of 3.
So, $567 \div 6 = 94$ with a remainder of 3.
This can also be written as a mixed number or a decimal.
As a mixed number: $94 \frac{3}{6} = 94 \frac{1}{2}$.
As a decimal: 94.5.

b) $\sqrt{916}$

$916 = 2 \times 458$
$458 = 2 \times 229$
So, $916 = 2 \times 2 \times 229 = 2^2 \times 229$.

The simplified form is $\boxed{2\sqrt{229}}$.

c) $\sqrt{891}$

$891 = 9 \times 99$
$9 = 3 \times 3$
$99 = 9 \times 11 = 3 \times 3 \times 11$
So, $891 = 3 \times 3 \times 3 \times 3 \times 11 = 3^4 \times 11$.

The simplified form is $\boxed{9\sqrt{11}}$.