how many 3 digit positive integers t exist that when divided by 7 leave a remainder of 5?  Answers.
A) 128  B) 142  C) 143  D) 141  E) 129
<gadget id="calculator">100 * 10</gadget>
<output>1_000</output>

<gadget id="calculator">1_000 - 1</gadget>
<output>999</output>

<gadget id="calculator">10 + 4</gadget>
<output>14</output>

<gadget id="calculator">14 * 7</gadget>
<output>98</output>

<gadget id="calculator">98 + 5</gadget>
<output>103</output>

<gadget id="calculator">999 - 103</gadget>
<output>896</output>

<gadget id="calculator">896 / 7</gadget>
<output>128</output>

<result>E</result>
how many 3 digit positive integers t exist that when divided by 7 leave a remainder of 5?
divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7)
"minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999 therefore , we identify the sequence 103 , 110,117 . . . . . 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the answer 129 that is definitely e ."