How many 3 digit positive integers T exist that when divided by 7 leave a remainder of 5? Choices:
A) 128	B) 142	C) 143	D) 141	E) 129
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= 
<gadget id="calculator">14 * 7</gadget>
<output>98</output>
98 +5= 
<gadget id="calculator">98 +5</gadget>
<output>103</output>
103   The second three digit number that leaves remainder 5 when divided by 7 is 110.   15 * 7= 
<gadget id="calculator">15 * 7</gadget>
<output>105</output>
105 +5= 
<gadget id="calculator">105 +5</gadget>
<output>110</output>
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= 
<gadget id="calculator">142 * 7</gadget>
<output>994</output>
994 + 5= 
<gadget id="calculator">994 + 5</gadget>
<output>999</output>
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 129.
<result>E</result>
How many 3 digit positive integers T exist that when divided by 7 leave a remainder of 5?