1. **State the problem:** We need to find four integers $a \leq b \leq c \leq d$ such that: - Their total sum is $a+b+c+d=44$. - The sum of the first two is $a+b=18$. - The sum of the middle two is $b+c=21$. - The sum of the last two is $c+d=26$. - The difference between the smallest and largest is $d - a = 7$. 2. **Use the given equations:** From $a+b=18$ and $b+c=21$, subtracting the first from the second gives: $$b+c - (a+b) = 21 - 18 \implies c - a = 3$$ From $c+d=26$ and $a+b=18$, adding these two gives: $$a+b+c+d = 18 + 26 = 44$$ which matches the total sum condition. 3. **Express variables in terms of $a$:** From $c - a = 3$, we get $c = a + 3$. From $d - a = 7$, we get $d = a + 7$. From $a + b = 18$, we get $b = 18 - a$. 4. **Check the middle two sum $b + c = 21$:** Substitute $b$ and $c$: $$b + c = (18 - a) + (a + 3) = 21$$ Simplify: $$18 - a + a + 3 = 21 \implies 21 = 21$$ This confirms the consistency. 5. **Find the values of $a, b, c, d$:** Since $a \leq b$, we have: $$a \leq 18 - a \implies 2a \leq 18 \implies a \leq 9$$ Since $b \leq c$, we have: $$18 - a \leq a + 3 \implies 18 - a \leq a + 3 \implies 18 - 3 \leq 2a \implies 15 \leq 2a \implies a \geq 7.5$$ Since $a$ is an integer, $a$ can be 8 or 9. 6. **Test $a=8$:** $$b = 18 - 8 = 10$$ $$c = 8 + 3 = 11$$ $$d = 8 + 7 = 15$$ Check order: $8 \leq 10 \leq 11 \leq 15$ correct. 7. **Test $a=9$:** $$b = 18 - 9 = 9$$ $$c = 9 + 3 = 12$$ $$d = 9 + 7 = 16$$ Check order: $9 \leq 9 \leq 12 \leq 16$ correct. 8. **Verify sums for both sets:** For $a=8$ set: $8+10+11+15=44$, sums match all conditions. For $a=9$ set: $9+9+12+16=46$ which is not 44, so discard. **Final answer:** The four integers are $\boxed{8, 10, 11, 15}$.